# Implicit Differentiation Tangent Line

Say you are building a ramp for a toy car, and you want to find the fastest possible ramp. After experimenting for a little while, you eventually conclude that the fastest possible ramp looks something like this:

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Fig 1. The fastest ramp between two points is called a brachistochrone curve (brachistochrone is Greek for “shortest time”)

Now, perhaps you decide to take a bike ride. Eventually, you start thinking about the curve traced out by the wheel of your bike. So, you decide to stick a piece of brightly colored paper on the wheel of your bike and plot its trajectory. You end up with a graph that looks something like this:

Fig. 2. The curve traced out by the wheel of a bicycle is a cycloid

To your surprise, this is the same shape you found when looking for the fastest ramp, just turned upside down. This shape is called a cycloid. After playing with the curve for a bit, you find the equation for the curve is given by

$x = \cos^{-1}(1-y)-\sqrt{y(2-y)}.$

Now, let's say you wanted to find the trajectory of the piece of paper or toy car at some point $$(x,y)$$. Normally, you would find the derivative of $$y$$ with respect to $$x$$, which would give you the slope of the tangent line to the curve at $$(x,y)$$. However, the equation above does not seem easy to solve in terms of $$y$$. What should you do?

This is a good situation in which to use implicit differentiation. This article will go over how to use implicit differentiation to implicitly find tangent lines to curves that do not necessarily have explicit formulas.

## Tangent Lines with Implicit Differentiation Meaning

Finding tangent lines with implicit differentiation just means applying implicit differentiation to find the slope of an implicitly defined curve. An implicit relation or implicitly defined curve is, in the context of calculus, an equation where the dependent variable is not isolated on one side. In other words, instead of being of the form

$y=f(x)$

for some function $$f$$, an implicit relation is of the form

$f(x,y) = g(x,y),$

where $$f$$ and $$g$$ are functions. The equation for the cycloid,

$x = \cos^{-1}(1-y)-\sqrt{y(2-y)},$

is an example of an implicit curve, since the variable $$y$$ is not isolated on one side of the equation. For more details and examples, see the article Implicit Relations.

Such equations can be difficult or even impossible to rewrite in an explicit form. So, to find their derivatives, you often need to use implicit differentiation. Implicit differentiation is an application of the chain rule to implicitly defined functions. For more details, see the article Implicit Differentiation.

Let's implicitly differentiate the relation

$y^5 + x^5 = \frac{\sqrt{\pi}}{17},$

assuming that $$y$$ is a function of $$x$$.

Solution:

• First, take the derivative of both sides of the equation with respect to $$x$$:

$\frac{d}{dx} (y^5 + x^5 ) = \frac{d}{dx} \left( \frac{\sqrt{\pi}}{17} \right).$

The right-hand side of the equation evaluates to zero since the derivative of a constant is always zero. To differentiate the left-hand side, recall that $$y$$ is assumed to be a function of $$x$$. So, let's write $$y = f(x)$$ for some function $$f$$:

$\frac{d}{dx} (y^5 + x^5 ) = \frac{d}{dx} \left((f(x))^5 + x^5 \right).$

• Since differentiation is linear, you can write:

$\frac{d}{dx} \left((f(x))^5 + x^5 \right) =\frac{d}{dx} (f(x))^5 + \frac{d}{dx} x^5 .$

• To differentiate the second term, use the Power Rule:

$\frac{d}{dx} (f(x))^5 + \frac{d}{dx} x^5 = \frac{d}{dx} (f(x))^5 + 5x^4.$

• To differentiate the first term, you need to use the Chain Rule, which gives you:

$\frac{d}{dx} (f(x))^5 + 5x^4 = 5(f(x))^4f'(x) + 5x^4.$

• So replacing $$f(x)$$ with $$y$$, and then setting the left and right-hand sides of the equation equal to each other, you get:

$\frac{d}{dx} (f(x))^5 + 5x^4 = 5(f(x))^4f'(x) + 5x^4 =0,$

or in other words

$5y^4 \frac{dy}{dx} + 5x^4 = 0.$

• Finally, you can solve for $$\frac{dy}{dx}$$ to get

$\frac{dy}{dx} = -\frac{5x^4}{5y^4} = - \frac{x^4}{y^4}.$

## Methods for Finding Tangent Lines with Implicit Differentiation

To find a tangent line at a point $$(x_1,y_1)$$ using implicit differentiation, you generally use the following method:

Step 1: Implicitly differentiate to find an expression for the derivative. This gives you the slope of the tangent line at any given point.

Step 2: Plug $$(x_1,y_1)$$ into the expression you found above to obtain the slope $$m$$ of the tangent line at $$(x_1,y_1)$$.

Step 3: Use the slope $$m$$ you obtained in Step 2 and the point $$(x_1,y_1)$$ to find the equation for the tangent line, using the formula $$y - y_1 = m(x-x_1)$$.

Let's take a closer look at each of these three steps, using the cycloid

$x = \cos^{-1}(1-y)-\sqrt{y(2-y)}$

at the point

$A = \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} , \frac{1}{2} \right)$

as an example.

Fig. 3. We can use implicit differentiation to find the equation of a tangent line to a cycloid.

## Using Implicit Differentiation to Find the Slope of the Tangent Line

First, you need to implicitly differentiate to find the slope of the tangent line. To accomplish this, first differentiate both sides of the relation with respect to $$x$$, using the Chain Rule where necessary. Then, plug in the point of interest and solve for the value of $$y'$$ at that point.

Let's implicitly differentiate the cycloid

$x = \cos^{-1}(1-y)-\sqrt{y(2-y)}$

to find the slope of the tangent line at the point

$A = \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} , \frac{1}{2} \right)$

Solution:

• Start by differentiating both sides of the equation with respect to $$x$$:

$\frac{d}{dx} x = \frac{d}{dx} \left(\cos^{-1}(1-y)-\sqrt{y(2-y)} \right).$

• You can use the Power Rule to find the derivative of the left-hand side:

$\frac{d}{dx} x = 1.$

• To differentiate the right-hand side, start by writing $$y = f(x)$$ for some function $$f$$: $\frac{d}{dx} \left(\cos^{-1}(1-y)-\sqrt{y(2-y)} \right)$ $= \frac{d}{dx} \left(\cos^{-1}(1-f(x))-\sqrt{f(x)(2-f(x))} \right) .$
• Since differentiation is linear, you can write:

$\frac{d}{dx} \left(\cos^{-1}(1-f(x))-\sqrt{f(x)(2-f(x))} \right)$ $= \frac{d}{dx}\left(\cos^{-1}(1-f(x))\right) - \frac{d}{dx} \left(\sqrt{f(x)(2-f(x))}\right).$

• You know that

$\cos^{-1}(x) = - \frac{1}{\sqrt{1-x^2}},$

so applying the Chain Rule, the first term becomes:

\begin{align} \frac{d}{dx}\left(\cos^{-1}(1-f(x))\right) & = -\frac{1}{\sqrt{1 - (1-f(x))^2}}(-f'(x)) \\ & = \frac{f'(x) }{\sqrt{1 - (1-f(x))^2}} . \end{align}

• Applying the Chain Rule, you can write the second term as:

$\frac{d}{dx} \left(\sqrt{f(x)(2-f(x))} \right)$ $= \left( \frac{1}{2} ((f(x) (2-f(x)))^{-\frac{1}{2}}\right)\left( \frac{d}{dx} f(x) (2-f(x)) \right).$

• Using that differentiation is linear, you can write:

\begin{align} \frac{d}{dx} f(x) (2-f(x)) &= \frac{d}{dx} \left( 2f(x) - (f(x))^2 \right) \\ &= 2\frac{d}{dx} f(x) - \frac{d}{dx}(f(x))^2 . \end{align}

• Using the Chain Rule, this expression becomes:

$2\frac{d}{dx} f(x) - \frac{d}{dx}(f(x))^2 = 2f'(x) -2f(x)f'(x).$

• That means the second term is actually

\begin{align} \frac{d}{dx} \left(\sqrt{f(x)(2-f(x))} \right) &= \left( \frac{1}{2} (f(x) (2-f(x)))^{-\frac{1}{2}}\right)\left( \frac{d}{dx} f(x) (2-f(x)) \right) \\ &= \left( \frac{1}{2} (f(x) (2-f(x)))^{-\frac{1}{2}}\right) \left( 2f'(x) -2f(x)f'(x) \right) \\ &= \frac{f'(x)(1-f(x))}{\sqrt{ f(x)(2-f(x))}}. \end{align}

• Combining the expressions for the first and second terms and rewriting in terms of $$y$$, the right-hand side of the equation is:

$\frac{f'(x) }{\sqrt{1 - (1-f(x))^2}} - \frac{f'(x)(1-f(x))}{\sqrt{ f(x)(2-f(x))}} = \frac{y' }{\sqrt{1 - (1-y)^2}} - \frac{y'(1-y)}{\sqrt{ y(2-y)}}.$

• Remember you already found out that

$\frac{d}{dx} \left(\cos^{-1}(1-y)-\sqrt{y(2-y)} \right) = 1.$

• Putting that together with the combined first and second terms,

$\frac{y' }{\sqrt{1 - (1-y)^2}} - \frac{y'(1-y)}{\sqrt{ y(2-y)}} =1 .$

At this point, you would usually solve for $$y'$$. However, you don't actually need to do that for a general point $$(x,y)$$ since you really only care about $$y'$$ at the point $$A$$.

• Substituting in $$A$$ gives you

$\frac{y' }{\sqrt{1 - \left(1-\frac{1}{2}\right)^2}} - \frac{y'\left(1-\frac{1}{2}\right)}{\sqrt{ \frac{1}{2} \left(2-\frac{1}{2} \right)}} =1 .$

• Now it is much easier to solve for $$y'$$. Doing a little simplifying first,

\begin{align} \frac{y' }{\sqrt{1 - \left(1-\frac{1}{2}\right)^2}} - \frac{y'\left(1-\frac{1}{2}\right)}{\sqrt{ \frac{1}{2} \left(2-\frac{1}{2} \right)}} &= \frac{y'}{\sqrt{\frac{3}{4}}} - \frac{\frac{1}{2}y'}{\sqrt{\frac{3}{4}}} \\ &= \frac{2}{\sqrt{3}}\left( y' - \frac{1}{2} y'\right) \\ &= \frac{y'}{\sqrt{3}}. \end{align}

So,

$\frac{y'}{\sqrt{3}} = 1,$

or

$y' = \sqrt{3},$

at the point $$A$$. In other words, the slope of the tangent line at $$A$$ is $$m = \sqrt{3}$$.

Once you are used to implicit differentiation, it can be faster and more efficient to skip the step of writing $$y$$ as $$f(x)$$. Instead, you can simply remember to multiply by $$y'$$ where necessary. This step is included to emphasize that, since $$y$$ is a function of $$x$$, implicit differentiation is really just an application of the Chain Rule.

Why does implicitly differentiating give you the slope of the tangent line? Well, implicit differentiation is just a way of finding the derivative of a curve at any given point. This is the same derivative you would get if differentiating normally, without using implicit differentiation. If you recall, the definition of the derivative of a function $$f$$ at a point $$x$$ is

$f'(x) = \lim\limits_{h \to 0} \frac{ f(x+h)-f(x)}{h}.$

Since

$\frac{ f(x+h)-f(x)}{h}$

is actually the slope of the line between the points $$(x+h, f(x+h))$$ and $$(x,f(x))$$, for very small $$h$$ the slope of the line should be very close to the slope of the tangent line at $$x$$. Remember that the derivative is the limit of this expression, and it is the exact slope of the tangent line. This will always be the case; it doesn't matter whether you find the derivative explicitly or implicitly.

## Using Implicit Differentiation to Find an Equation for the Tangent Line

Once you have found the slope $$m$$ of the tangent line at the point $$(x_1,y_1)$$, all you need to do is plug the values you found into the formula $$y - y_1 = m(x-x_1)$$ and simplify the expression.

But is $$(x_1,y_1)$$ actually a point on the tangent line? If you set $$x = x_1$$ in the equation for the tangent line, you get that $$y - y_1 = 0$$, so $$y=y_1$$. Thus, the point $$(x_1,y_1)$$ is indeed a point on the tangent line. By definition, this line also has slope $$m$$, so it is the tangent line you are looking for.

Now that you have found the slope of the cycloid

$x = \cos^{-1}(1-y)-\sqrt{y(2-y)}$

at the point

$A = \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} , \frac{1}{2} \right) ,$ you can use it to find the equation for the tangent line at $$A$$.

Solution:

In the previous example you found that the slope of the tangent line at $$A$$ is $$m = \sqrt{3}$$. Then plugging $$m$$ and $$A$$ into the equation for the tangent line,

$y - \frac{1}{2} = \sqrt{3} \left( x - \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) \right).$

If you simplify this you get the expression

\begin{align} y &= x\sqrt{3} - \frac{\pi\sqrt{3}}{3} + \frac{3}{2} + \frac{1}{2} \\ &= \sqrt{3}x + \frac{6-\pi\sqrt{3}}{3}.\end{align}

Graphing this line, you see that this is indeed the equation for the tangent line to the cycloid at $$A$$.

Fig. 4. Implicit differentiation can be used to find the equation of the tangent line to the graph of a cycloid.

## Finding Tangent Lines with Implicit Differentiation Examples

Let's look at an example to practice finding tangent lines with implicit differentiation.

Let's find the equation of the tangent line to the curve

$y^2 (3x^2 -y^2)^2 = (x^2 + y^2)^4$

at

$A = \left( \frac{1}{2}, \frac{1}{2} \right).$

Fig. 5. A rose curve with six petals

First, differentiate both sides of the equation:

$\frac{d}{dx} \left(y^2 (3x^2 -y^2)^2 \right) = \frac{d}{dx} (x^2 + y^2)^4.$

Since $$y$$ is a function of $$x$$, you can write $$y=f(x)$$ for some function $$f$$:

$\frac{d}{dx} \left((f(x))^2 (3x^2 -(f(x))^2)^2 \right) = \frac{d}{dx} (x^2 + (f(x))^2)^4.$

The right-hand side of the equation can be differentiated using the Chain Rule to give you:

\begin{align} \frac{d}{dx} (x^2 + (f(x))^2)^4 &= 4(x^2 + (f(x))^2)^3\left( \frac{d}{dx} \left( (x^2 + (f(x))^2\right) \right) \\ &= 4(x^2 + (f(x))^2)^3\left( 2x + 2f(x)f'(x) \right) . \end{align}

To differentiate the left-hand side of the equation, use the Product Rule and the Chain Rule to get:

\begin{align} \frac{d}{dx} \left((f(x))^2 (3x^2 -(f(x))^2)^2 \right) &= \left(\frac{d}{dx}(f(x))^2\right)\left( (3x^2 - (f(x))^2 \right)^2 \\&\quad + (f(x))^2\frac{d}{dx}\left( (3x^2 - (f(x))^2)^2\right) \\&= 2f(x)f'(x) ((3x^2 - (f(x))^2)^2 \\&\quad + (f(x))^2 \cdot 2(3x^2 - (f(x))^2) \frac{d}{dx} \left( (3x^2 - (f(x))^2\right) \\&= 2f(x)f'(x) (3x^2 - (f(x))^2)^2 \\&\quad + 2 (f(x))^2 (3x^2 - (f(x))^2) (6x - 2f(x)f'(x)).\end{align}

Combining the right and left-hand sides of the equation and replacing $$f(x)$$ with $$y$$ gives you:

$2yy'(3x^2 - y^2)^2 + 2y^2(3x^2-y^2)(6x-2yy') = 4(x^2+y^2)^3(2x+2yy').$

Now plug the point $$A$$ into the equation above and solve for $$y'$$. Plugging $$A$$ into the left-hand side of the equation gives:

\begin{align} & \left. 2yy'(3x^2 - y^2)^2 + 2y^2(3x^2-y^2)(6x-2yy') \right|_A \\ &\quad = 2\cdot \frac{1}{2}y'\left(3\left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) ^2\right)^2 \\&\quad\quad + 2\left(\frac{1}{2}\right) ^2\left(3\left(\frac{1}{2}\right) ^2-\left(\frac{1}{2}\right) ^2\right)\left(6\left(\frac{1}{2}\right) -2\left(\frac{1}{2}\right) y'\right) \\&\quad= y'\left(\frac{2}{4}\right)^2 + \frac{2}{4}\cdot\frac{2}{4}(3-y') \\&\quad= \frac{y'}{4} + \frac{3}{4} - \frac{y'}{4} \\&\quad= \frac{3}{4}. \end{align}

Plugging $$A$$ into the right-hand side of the equation gives the expression:

\begin{align} \left. 4(x^2+y^2)^3(2x+2yy') \right|_A &= 4\left(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right) ^2\right)^3\left(2\cdot\left(\frac{1}{2}\right) +2\left(\frac{1}{2}\right) y'\right) \\&= \frac{4}{8} + \frac{4y'}{8} \\ &= \frac{1}{2} + \frac{y'}{2}. \end{align}

Setting the two sides equal to each other gives

$\frac{3}{4} =\frac{1}{2} + \frac{y'}{2} ,$

or in other words

$y' = \frac{1}{2}.$

Finally, plug this value of $$y'$$ and the point $$A$$ into the equation for the tangent line at a point, giving you the equation:

$y - \frac{1}{2} = \frac{1}{2} \left( x - \frac{1}{2} \right) .$

Solving for $$y$$ gives you the tangent line equation

$y = \frac{1}{2}x + \frac{1}{4}.$

Plotting this line, you see that this is indeed the line you are looking for.

Fig. 6. This curve is called a rose curve, one of a family of curves that look like flowers.

## Using Implicit Differentiation to Find Equations of Normal Lines

A normal line to a curve at a point is the line perpendicular to the tangent line at that point.

Fig. 7. The normal to a curve is the line perpendicular to the tangent line.

Normal lines often show up in physics. For example, the force of friction is normal to the curve describing an object's trajectory. As another example, Snell's law, an important law in optics that is also called the law of refraction, is formulated in terms of normal lines.

Remember the brachistochrone curve from earlier, the one that gave us the fastest toy car ramp? Believe it or not, one of the ways to prove that the brachistochrone is indeed the fastest ramp involves Snell's law! This proof was discovered by Johann Bernoulli and works by imagining that the toy car is a light ray traveling through mediums of different densities.1

To find the normal to a curve at a point $$A = (x_1, y_1)$$, first, differentiate to find the slope $$m$$ of the tangent line at $$A$$. You might remember from geometry that the slopes of perpendicular lines are negative reciprocals of each other. Since the normal line is perpendicular to the tangent line and the tangent has slope $$m$$, the slope of the normal line is

$-\frac{1}{m}.$

Finally, the equation for the normal line at $$A$$ is

$y - y_1 = -\frac{1}{m}(x-x_1).$

Let's do an example of finding the normal line to a curve using implicit differentiation.

Let's find the equation of the normal line to the curve $$(x^2+y^2)^2=2(x^2-y^2)+\tfrac{1}{4}$$ at $$\left( \tfrac{1}{2},\tfrac{1}{2}\right)$$

• First, we differentiate both sides of the equation with respect to x:

$\frac{\mathrm{d}}{\mathrm{d}x} \left((x^2+y^2)^2\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(2(x^2-y^2)+\dfrac{1}{4}\right)$

• Using the Chain Rule, the left-hand side of the equation becomes:

\begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \left((x^2+y^2)^2\right) &=2(x^2+y^2)\frac{\mathrm{d}}{\mathrm{d}x}(x^2+y^2) \\ &=2(x^2+y^2)(2x+2yy')\end{align}

Since $$y$$ is a function of $$x$$, remember to indicate its derivative as $$y'$$.

• Using the Chain Rule and Power Rule, the right-hand side of the equation becomes:

\begin{align} \frac{\mathrm{d}}{\mathrm{d}x} \left(2(x^2-y^2)+\dfrac{1}{4}\right) &= 2\frac{\mathrm{d}}{\mathrm{d}x} \left(x^2-y^2\right)+0 \\ &=2(2x-2yy')\end{align}

• Combining both expressions, we get:

$2(x^2+y^2)(2x+2yy')=2(2x-2yy')$

• Next, we plug the point $$\left( \tfrac{1}{2},\tfrac{1}{2}\right)$$ into the expression above to find the slope of the tangent line at that point. Doing so, we get:

$2\left(\left( \dfrac{1}{2}\right)^2+\left( \dfrac{1}{2}\right)^2\right)\left(2\dfrac{1}{2}+2\dfrac{1}{2}y'\right)=2\left(2\dfrac{1}{2}-2\dfrac{1}{2}y'\right)$

• Simplifying, this gives the expression:

$1+y'=2-2y'$

• Solving for y', we get:

$y'=\dfrac{1}{3}$

So, the slope of the tangent line at $$\left( \tfrac{1}{2},\tfrac{1}{2}\right)$$ is $$\tfrac{1}{3}$$. To obtain the slope of the normal line, we take the negative reciprocal of the slope of the tangent line, getting $$-3$$.

Finally, we can plug our point and slope into the equation for the normal line, giving us:

$y-\dfrac{1}{2}=-3\left(x-\dfrac{1}{2}\right)$

Simplifying and solving for $$y$$, we get that the equation for the normal line (graphed below) is

$y=-3x+2$.

Fig. 8. The normal line to an implicit curve is perpendicular to the tangent line and can be found using implicit differentiation

## Implicit Differentiation Tangent Line - Key takeaways

• Implicit differentiation is used to find tangent lines to implicitly defined curves.
• The equation of a line with slope $$m$$ through a point $$(a,b)$$ is $$y - b = m(x - a)$$.
• Use the derivative to find the slope of the curve at $$A$$, then solve for the equation of the tangent line using this slope and the point $$A$$.
• The normal line to a curve is the line perpendicular to the tangent line.

## References

1. John Martin, “The Helen of Geometry,” The College Mathematics Journal 41, no. 1 (January 2010).

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How do you use implicit differentiation to find an equation of the tangent line?

To find the equation of a tangent line using implicit differentiation, first implicitly differentiate to find the derivative of the curve. Then, use the equation for a line with known slope passing through a point to find the equation for the tangent line.

How to use implicit differentiation to find the slope of a tangent line?

First, implicitly differentiate to find an expression for the derivative of the curve. The slope of a tangent line at any point on the curve is the same as the value of the derivative at that point.

What is an example of finding a tangent line with implicit differentiation?

An example where implicit differentiation can be used to find tangent lines is finding the tangent line to the unit circle at the point (0,1).

What is the formula for implicit differentiation?

Implicit differentiation is an application of the Chain Rule to implicitly defined curves. To implicitly differentiate, differentiate each side of the relation as usual, then multiply each y term by y'.

Is the gradient of the tangent equal to the derivative?

The gradient of a tangent line to a curve at a point is equal to the derivative of the curve at that point.

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