Implicit Differentiation

Implicit differentiation is an approach to taking derivatives that uses the chain rule to avoid solving explicitly for one of the variables. For example, if \( y + 3x = 8, \) we can directly take the derivative of each term with respect to \(x\) to obtain \(\frac{dy}{dx} + 3 = 0,\) so \(\frac{dy}{dx} = -3.\)

Implicit differentiation is especially useful where it is difficult to isolate one of the variables in the given relationship. For example, if \(y = x^2 + y^2,\) solving for \(y\) and then taking the derivative would be painful. Instead, using implicit differentiation to directly take the derivative with respect to \(x\) gives

\[\frac{dy}{dx} = 2x + 2y\frac{dy}{dx},\]

so \(\frac{dy}{dx} = \frac{2x}{1-2y}.\) This can be especially useful in fields like physics where relationships between variables are often known but a nice closed-form is not attainable.

Given \( x^2 + x + y^2 = 15 \), what is \( \frac{dy}{dx} \) at the point \( ( 2, 3)?\)

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Since \( x^2 + x + y^2 = 15 \), differentiating, we have

\[ \begin{align} 2x+1+2y\left(\frac{dy}{dx}\right)&=0 \\ \frac{dy}{dx}&=-\frac{2x+1}{2y}. \end{align}\]

Thus \( \frac{dy}{dx} \) at the point \( ( 2, 3) \) is \( -\frac{2(2)+1}{2(3)}=-\frac{5}{6} \). \( _\square \)

Alternatively, we can use the fact that if \(R(x,y)=0\), then \(\frac{dy}{dx}=-\frac{\hspace{3mm} \frac{\partial R}{\partial x}\hspace{3mm} }{\frac{\partial R}{\partial y}}.\) Rewriting the function,

\[x^2 + x + y^2 - 15 = 0.\]

Applying the method given above,

\[ \begin{align} \frac{dy}{dx} &= - \frac{\frac{\partial}{\partial x}( x^2 + x + y^2 - 15)}{\frac{\partial}{\partial y}( x^2 + x + y^2 - 15)} \\\\ &= - \dfrac{2x +1}{2y}. \end{align}\]

Thus \( \frac{dy}{dx} \) at the point \( ( 2, 3) \) is \( -\frac{2(2)+1}{2(3)}=-\frac{5}{6} \). \( _\square \)

Implicit differentiation is a technique that can be used to differentiate equations that are not given in the form of \(y=f(x).\) For instance, the differentiation of \(x^2+y^2=1\) looks pretty tough to do by using the differentiation techniques we've learned so far (which were explicit differentiation techniques), since it is not given in the form of \(y=f(x).\) Observe that the term equations was used in the first sentence. In fact, implicit differentiation can be used to differentiate equations that do not match the criteria to be a function.

The main key to implicit differentiation is to keep in mind with respect to which variable we are differentiating. Here is an easy explanation of how it is done:

When differentiating an equation implicitly with respect to \(x,\) we should

(1) add \(\frac{d}{dx}\) to all of the terms;
(2) apply the iterated chain rule to any terms that are not expressed in terms of \(x.\)

The first step is the process of differentiating both sides of an equation with respect to \(x.\) The second step gives the solution to terms that are not described by \(x.\) Remember the iterated chain rule? Here is a quick reminder:

\[\frac{dz}{dx}=\frac{dz}{dy}\cdot\frac{dy}{dx}.\]

Implicit differentiation can be used in differentiating rational functions for convenience. Although most rational functions are given in the form of \(y=f(x),\) differentiating these explicitly would require using the quotient rule, which is pretty annoying. A slight modification to the equation (which in most cases would be multiplying the quotient to both sides) and the application of implicit differentiation would ease up obtaining the derivative. Here are some examples:

Find the derivative of \(y=\frac{1}{x+1}\) using implicit differentiation.

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To avoid using the quotient rule, we can change both sides to their reciprocals:

\[\frac{1}{y}=x+1.\]

Then we differentiate both sides:

\[\begin{align} \frac{d}{dx}\frac{1}{y}&=\frac{d}{dx}(x+1)\\ \frac{d}{dx}\frac{1}{y}&=1, \end{align}\]

and apply the chain rule:

\[\begin{align} \frac{d}{dx}\frac{1}{y}&=1\\ \frac{d}{dy}\frac{1}{y}\cdot\frac{dy}{dx}&=1\\ -\frac{1}{y^2}\cdot\frac{dy}{dx}&=1\\ \Rightarrow \frac{dy}{dx}&=-y^2\\&=-\frac{1}{(x+1)^2}.\ _\square \end{align}\]

Find the derivative of \(y=\frac{2x-1}{3x+2}\) using implicit differentiation.

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First, we multiply both sides by \(3x+2\) to get rid of the quotient:

\[y(3x+2)=2x-1.\]

Then we differentiate both sides of the equation. Obviously, the product rule should be used for the left-hand side:

\[\begin{align} y(3x+2)&=2x-1\\ \frac{d}{dx}\big(y(3x+2)\big)&=\frac{d}{dx}(2x-1)\\ \left(\frac{d}{dx}y\right)\cdot(3x+2)+3y&=2\\ \frac{dy}{dx}&=\frac{2-3y}{3x+2}. \end{align}\]

If you don't like having the term \(y\) in the answer, you can plug in \(y=\frac{2x-1}{3x+2},\) which gives

\[\frac{dy}{dx}=\frac{2x-3y-1}{3x+2}=\frac{2x-3\cdot\frac{2x-1}{3x+2}-1}{3x+2}=\frac{7}{(3x+2)^2}.\]

However, this procedure isn't necessary since we can always find the value of \(y\) when given the value of \(x.\)

Of course, using explicit differentiation by applying the quotient rule will also lead to the same answer. Try it out for yourself. \(_\square\)

Find the derivative of \(y=\frac{x^3+2}{x^2-2x-4}\) using implicit differentiation.

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We have

\[\begin{align} y&=\frac{x^3+2}{x^2-2x-4}\\ (x^2-2x-4)y&=x^3+2\\ \frac{d}{dx}\big((x^2-2x-4)y\big)&=\frac{d}{dx}(x^3+2)\\ (2x-2)y+(x^2-2x-4)\cdot\frac{dy}{dx}&=3x^2\\ \Rightarrow \frac{dy}{dx}&=\frac{3x^2-(2x-2)y}{x^2-2x-4}.\ _\square \end{align}\]

Find the tangent to the curve \(y=\frac{3x-7}{x^2-4x+1}\) at \(x=0.\)

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First, we should find the slope of the tangent, which is equal to \(\left.\frac{dy}{dx}\right|_{x=0}.\) We have

\[\begin{align} y&=\frac{3x-7}{x^2-4x+1}\\ (x^2-4x+1)y&=3x-7\\ (2x-4)y+(x^2-4x+1)\cdot\frac{dy}{dx}&=3\\ \Rightarrow \frac{dy}{dx}&=\frac{3-(2x-4)y}{x^2-4x+1}. \end{align}\]

Since \(\left. y\right|_{x=0}=-7,\) we have

\[\left.\frac{dy}{dx}\right|_{x=0}=\frac{3-28}{1}=-25.\]

Therefore, the equation of the tangent is

\[\begin{align} y&=-25(x-0)-7\\ y&=-25x-7.\ _\square \end{align}\]

Find the derivative of \(x^2+y^2=1.\)

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First, we add \(\frac{d}{dx}\) to both sides of the equation, which gives

\[\begin{align} \frac{d}{dx}x^2+\frac{d}{dx}y^2&=\frac{d}{dx}1\\ 2x+\frac{d}{dx}y^2&=0. \end{align}\]

Then we use the chain rule to solve \(\frac{d}{dx}y^2:\)

\[\begin{align} 2x+\frac{d}{dx}y^2&=0\\ 2x+\frac{d}{dy}y^2\cdot\frac{dy}{dx}&=0\\ 2x+2y\cdot\frac{dy}{dx}&=0\\ \Rightarrow \frac{dy}{dx}&=-\frac{x}{y}. \end{align}\]

Notice that the answer can contain variables other than \(x,\) for it would be messy to express the equation using only one variable. \(_\square\)

Find the derivative of \(2y^3-y^2=x^5+3x^2.\)

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We have

\[\begin{align} \frac{d}{dx}2y^3-\frac{d}{dx}y^2&=\frac{d}{dx}x^5+\frac{d}{dx}3x^2\\ \frac{d}{dy}2y^3\cdot\frac{dy}{dx}-\frac{d}{dy}y^2\cdot\frac{dy}{dx}&=5x^4+6x\\ \frac{dy}{dx}\cdot(6y^2-2y)&=5x^4+6x\\ \Rightarrow \frac{dy}{dx}&=\frac{5x^4+6x}{6y^2-2y}.\ _\square \end{align}\]

Find the derivative of \(\ln y+e^y=\sin y^2-3\cos x.\)

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We have

\[\begin{align} \frac{d}{dx}\ln y+\frac{d}{dx}e^y&=\frac{d}{dx}\sin y^2-\frac{d}{dx}3\cos x\\ \frac{d}{dy}\ln y\cdot\frac{dy}{dx}+\frac{d}{dy}e^y\cdot\frac{dy}{dx}&=\frac{d}{dy}\sin y^2\cdot\frac{dy}{dx}+3\sin x\\ \frac{dy}{dx}\left(\frac{1}{y}+e^y-2y\cos y^2\right)&=3\sin x\\ \Rightarrow \frac{dy}{dx}&=\frac{3\sin x}{\frac{1}{y}+e^y-2y\cos y^2}.\ _\square \end{align}\]

Find the tangent to the curve \(\sqrt{x}+\sqrt{y}=1\) at \(\left(\frac{1}{4},\frac{1}{4}\right).\)

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First, we find the slope which is equal to \(\left.\frac{dy}{dx}\right|_{x=\frac{1}{4}}.\) Differentiating both sides with respect to \(x\) gives

\[\begin{align} \frac{d}{dx}\sqrt{x}+\frac{d}{dx}\sqrt{y}&=\frac{d}{dx}1\\ \frac{1}{2\sqrt{x}}+\frac{d}{dy}\sqrt{y}\cdot\frac{dy}{dx}&=0\\ \frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}\cdot\frac{dy}{dx}&=0\\ \frac{dy}{dx}&=-\frac{\sqrt{y}}{\sqrt{x}}. \end{align}\]

Therefore the slope of the tangent is

\[\left.\frac{dy}{dx}\right|_{x=\frac{1}{4}}=\left.-\frac{\sqrt{y}}{\sqrt{x}}\right|_{x=\frac{1}{4}}=-1.\]

Since the tangent passes through the point \(\left(\frac{1}{4},\frac{1}{4}\right),\) the equation of the tangent is

\[\begin{align} y&=-\left(x-\frac{1}{4}\right)+\frac{1}{4}\\ &=-x+\frac{1}{2}.\ _\square \end{align}\]

Find \(\frac{dy}{dx}\) if \(y^2=x^2+\sin(xy).\)

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Using implicit differentiation,

\[\begin{align} \dfrac{d}{dx}\big(y^2\big)&=\dfrac{d}{dx}\big(x^2\big)+\dfrac{d}{dx}(\sin xy)\\ 2y\dfrac{dy}{dx}&=\dfrac{d}{dx}\big(x^2\big)+(\cos xy)\dfrac{d}{dx}(xy)\\ 2y\dfrac{dy}{dx}&=2x+(\cos xy)\left(y+x\dfrac{dy}{dx}\right)\\ 2y\dfrac{dy}{dx}-(\cos xy)\left(x\dfrac{dy}{dx}\right)&=2x+\cos (xy)y\\ (2y-x\cos xy)\dfrac{dy}{dx}&=2x+\cos (xy)y\\ \dfrac{dy}{dx}&=\dfrac{2x+\cos (xy)y}{2y-x\cos xy}.\ _\square \end{align}\]

We can also use implicit differentiation to find the derivatives of the inverse trigonometric functions, assuming that these functions are differentiable.

Find the derivative of \(y=\sin^{-1} x\) using implicit differentiation.

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Since \(y=\sin^{-1} (x),\) we have \(x=\sin y,\) where \(- \frac{\pi}{2} \leq y \leq\frac{\pi}{2}.\)

Differentiating implicitly, we have

\[\cos y \dfrac{dy}{dx}=1 \implies \dfrac{dy}{dx}=\dfrac{1}{\cos y}.\]

Since \(-\frac{\pi}{2} \leq y \leq\frac{\pi}{2}\) and \(\cos y \geq 0,\) using \(\cos y=\sqrt{1-\sin^2 y}=\sqrt{1-x^2},\)

we have \(\frac{d}{dx} \big(\sin^{-1} x\big) = \frac{1}{\sqrt{1-x^2}}.\) \(_\square\)

Implicit differentiation is an approach to taking derivatives that uses the chain rule to avoid solving explicitly for one of the variables.

If \( y^6 - e^{xy} = x, \) what is \( \frac{dy}{dx}?\)

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Taking the derivative of every term with respect to \( x \) gives us

\[ \begin{align} 6y^5\left( \frac{dy}{dx} \right) - e^{xy} \left(\frac{d}{dx} (xy)\right)&=1 \\ 6y^5\left( \frac{dy}{dx} \right) - e^{xy} \left( y + x\frac{dy}{dx} \right)&=1 . \end{align}\]

Now, we can isolate all of the \( \frac{dy}{dx} \) on the left:

\[ \begin{align} 6y^5\left( \frac{dy}{dx} \right) - xe^{xy}\left(\frac{dy}{dx}\right) &=1 + ye^{xy} \\ \left( \frac{dy}{dx} \right) \left(6y^5 - xe^{xy} \right)&=1 + ye^{xy} \\ \frac{dy}{dx} &= \frac{ 1 + ye^{xy} }{ 6y^5 - xe^{xy} }. \ _\square \end{align}\]

The area of a circle is given by \(A=\pi r^2.\) If the radius and area are differentiable functions of time \(t,\) find the equation relating \(\frac{dA}{dt}\) and \(\frac{dr}{dt}.\)

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Let's differentiate the expression \(A=\pi r^2\) with respect to \(t,\) so we get:

\[\begin{align} \dfrac{dA}{dt}&=\pi \dfrac{dr}{dt}(2r)\\ &=2\pi r\dfrac{dr}{dt}.\ _\square \end{align}\]