Implicit and Logarithmic Differentiation

In our work up until now, the functions we needed to differentiate were either given explicitly, such as $\mathrm{y}=\mathrm{f}(\mathrm{x})=$ $x^{2}+\sin (x)$, or it was possible to get an explicit formula for them, such as solving

$y^{3}-3 x^{2}=5$ to get $y=\sqrt[3]{5+3 x^{2}}$. Sometimes, however, we will have an equation relating $x$ and $y$ which is either difficult or impossible to solve explicitly for $y$, such as $y^{2}+2 y=\sin (x)+4$ or $y+\sin (y)=x^{3}-x.$ In any case, we can still find $y^{\prime}=f^{\prime}(x)$ by using implicit differentiation.

The key idea behind implicit differentiation is to assume that $\mathbf{y}$ is a function of $\mathbf{x}$ even if we cannot explicitly solve for $\mathrm{y}$. This assumption does not require any work, but we need to be very careful to treat $\mathrm{y}$ as a function when we differentiate and to use the Chain Rule or the Power Rule for Functions.

Example 1: Assume that $\mathrm{y}$ is a function of $\mathrm{x}$.

Calculate (a) $\mathbf{D}\left(\mathrm{y}^{3}\right)$

(b) $\frac{\mathbf{d}}{\mathbf{d x}}\left(x^{3} y^{2}\right)$, and

(c) $(\sin (\mathrm{y}))^{\prime}$

Solution: (a) We need the Power Rule for Functions since $\mathbf{y}$ is a function of $\mathbf{x}$:

$\mathbf{D}\left(\mathrm{y}^{3}\right)=3 \mathrm{y}^{2} \cdot \mathbf{D}(\mathrm{y})=3 \mathrm{y}^{2} \cdot \mathbf{y}^{\prime}$

(b) We need to use the product rule and the Chain Rule:

$\frac{d}{d x}\left(x^{3} y^{2}\right)=x^{3} \cdot \frac{d}{d x}\left(y^{2}\right)+y^{2} \cdot \frac{d}{d x}\left(x^{3}\right)=x^{3} 2 y \cdot \frac{d y}{d x}+y^{2} \cdot 3 x^{2}=2 x^{3} y \cdot \frac{d y}{d x}+3 x^{2} y^{2}$

(c) We just need to know that $\mathbf{D}(\sin (x))=\cos (x)$ and then use the Chain Rule:

$(\sin (\mathrm{y}))^{\prime}=\cos (\mathrm{y}) \cdot \mathbf{y}^{\prime}$

Practice 1: Assume that $\mathrm{y}$ is a function of $x$. Calculate (a) $\mathbf{D}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)$ and (b) $\frac{\mathbf{d}}{\mathbf{d x}}(\sin (2+3 \mathrm{y}))$.

Example 2: Find the slope of the tangent line to the circle $\mathrm{x}^{2}+\mathrm{y}^{2}=25$ at the point $(3,4)$ with and without implicit differentiation.

Solution:

Explicitly: We can solve the equation of the circle for $\mathrm{y}=+\sqrt{25-\mathrm{x}^{2}}$ or $\mathrm{y}=-\sqrt{25-\mathrm{x}^{2}}$.

Since the point $(3,4)$ is on the top half of the circle (Fig. 1), $\mathrm{y}=+\sqrt{25-\mathrm{x}^{2}}$ and

$\mathbf{D}(\mathrm{y})=\mathbf{D}\left(+\sqrt{25-\mathrm{x}^{2}}\right)=\frac{1}{2}\left(25-\mathrm{x}^{2}\right)^{-1 / 2} \mathbf{D}\left(25-\mathrm{x}^{2}\right)=\frac{-\mathrm{x}}{\sqrt{25-\mathrm{x}^{2}}}$

Replacing $x$ with 3, we have $y^{\prime}=\frac{-3}{\sqrt{25-3^{2}}}=-3 / 4$.

Implicitly: We differentiate each side of the equation $\mathrm{x}^{2}+\mathrm{y}^{2}=25$ and then solve for $\mathbf{y}^{\prime} \cdot \mathbf{D}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)=\mathbf{D}(25)$ so $2 \mathrm{x}+2 \mathrm{y} \cdot \mathbf{y}^{\prime}=0$.

Solving for $\mathbf{y}^{\prime}$, we have $\mathbf{y}^{\prime}=-\frac{2 \mathrm{x}}{2 \mathrm{y}}=-\mathrm{x} / \mathrm{y}$, and, at the point $(3,4)$, $\mathbf{y}^{\prime}=-3 / 4$, the same answer we found explicitly.

Practice 2: Find the slope of the tangent line to $\mathrm{y}^{3}-3 \mathrm{x}^{2}=15$ at the point $(2,3)$ with and without implicit differentiation.

In the previous example and practice problem, it was easy to explicitly solve for $\mathrm{y}$, and then we could differentiate $y$ to get $\mathrm{y}^{\prime}$. Because we could explicitly solve for $\mathrm{y}$, we had a choice of methods for calculating $\mathrm{y}^{\prime}$. Sometimes, however, we can not explicitly solve for $\mathrm{y}^{\prime}$, and the only way of determining $\mathrm{y}^{\prime}$ is implicit differentiation.

Example 3: Determine $\mathrm{y}^{\prime}$ at $(0,2)$ for $y^{2}+2 y=\sin (x)+8$.

Solution: Assuming that $\mathrm{y}$ is a function of $\mathrm{x}$ and differentiating each side of the equation, we get

$\mathbf{D}\left(\mathrm{y}^{2}+2 \mathrm{y}\right)=\mathrm{D}(\sin (\mathrm{x})+8) \quad$ so $2 \mathrm{y} \mathbf{y}^{\prime}+2 \mathbf{y}^{\prime}=\cos (\mathrm{x}) \quad$ and $(2 \mathrm{y}+2) \mathbf{y}^{\prime}=\cos (\mathrm{x})$

Then $\mathbf{y}^{\prime}=\frac{\cos (\mathrm{x})}{2 \mathrm{y}+2} \quad$ so, at the point $(0,2), \mathbf{y}^{\prime}=\frac{\cos (0)}{2(\mathbf{2})+2}=1 / 6$.

Practice 3: Determine $\mathrm{y}^{\prime}$ at $(1,0)$ for $y+\sin (y)=x^{3}-x$.

In practice, the equations may be rather complicated, but if you proceed carefully and step-by-step, implicit differentiation is not difficult. Just remember that $\mathrm{y}$ must be treated as a function so every time you differentiate a term containing a $\mathrm{y}^{\prime}$ you should get something which has a $\mathbf{y}^{\prime}.$ The algebra needed to solve for $\mathrm{y}^{\prime}$ is always easy - if you differentiated correctly the resulting equation will be a linear equation in the variable $\mathrm{y}^{\prime}$.

Example 4: Find the equation of the tangent line $\mathrm{L}$ to the "tilted' parabola in Fig. 1 at the point $(1,2)$.

Solution: The line goes through the point $(1,2)$ so we need to find the slope there. Differentiating each side of the equation, we get

$\mathbf{D}\left(x^{2}+2 x y+y^{2}+3 x-7 y+2\right)=D(0)$ so

$2 x+2 x y^{\prime}+2 y+2 y y^{\prime}+3-7 y^{\prime}=0$ and

$(2 x+2 y-7) y^{\prime}=-2 x-2 y-3$.

Solving for $y^{\prime}, y^{\prime}=\frac{-2 x-2 y-3}{2 x+2 y-7}$, so the slope at $(1,2)$ is $m=y^{\prime}=\frac{-2-4-3}{2+4-7}=9$.

Finally, the equation of the line is $y-2=9(x-1)$ so $y=9 x-7$.

Practice 4: Find the points where the graph in Fig. 2 crosses the y-axis, and find the slopes of the tangent lines at those points.

Implicit differentiation is an alternate method for differentiating equations which can be solved explicitly for the function we want, and it is the only method for finding the derivative of a function which we cannot describe explicitly.