Implicit Differentiation Tangent Line

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

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

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 \(m = \sqrt{3}\).

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.

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.

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