Implicit Differentiation:

Suppose we are given the following equation y³ -7y² + 4x²= e^x -4. Can we rewrite this equatiion in the form y = something in terms of x? Not in this case. So how do we determine how fast y changes as x changes?

How do we find $$\frac{dy}{dx}$$?

We need to differentiate each component of the equation with respect to x. Each of the terms just in x can be differentiated as normal. But the terms with y's must be differentiated with the chain rule. Notice that each term with y's is essentially a function of y. Remember the Chain Rule where u=y³ is some f(y). Then

$$\displaystyle\frac{d}{dx}\left[y^3\right] = \displaystyle\fra... ...u}{dy}\displaystyle\frac{dy}{dx}=3y^2\displaystyle\frac{dy}{dx}$$

Using this method for each component we get the following:

$$3y^2\left(\displaystyle\frac{dy}{dx}\right)-14y\left(\displaystyle\frac{dy}{dx}\right)+8x=e^x$$

Now group all of the $$\frac{dy}{dx}$$ terms together:

$$(3y^2-14y)\left(\displaystyle\frac{dy}{dx}\right)=e^x -8x$$

If follows that

$$\displaystyle\frac{dy}{dx}=\displaystyle\frac{e^x -8x}{3y^2-14y}$$

Now for any point (x,y) on our graph we have an equation for the slope of the tangent line to the graph at that point.


We can also use implicit differentiation to solve more complicated explicit functions such as

$$y=\displaystyle\frac{(x-1)^{3/2}}{\sqrt{x+1}}$$

First take the logarithm of both sides.

$$\ln(y)=\ln\left(\displaystyle\frac{(x-1)^{3/2}}{\sqrt{x+1}}\right)$$

$$\ln(y)=\ln((x-1)^{3/2})-\ln(\sqrt{x+1})$$

$$\ln(y)=\displaystyle\frac{3}{2}\ln(x-1)-\ln(\sqrt{x+1})$$

Next differentiate,

$$\left(\displaystyle\frac{1}{y}\right)\left(\displaystyle\frac... ...yle\frac{1}{\sqrt{x+1}}\right)\displaystyle\frac{1}{\sqrt{x+1}}$$

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

$$\displaystyle\frac{dy}{dx}=\displaystyle\frac{3y}{2(x-1)}-\displaystyle\frac{y}{2(x+1)}$$

$$\displaystyle\frac{dy}{dx}=\displaystyle\frac{y+2yx+3}{2(x^2-1)}$$

Even though this looks comlex, imagine using the quotient rule on the previous equation. Here we took the derivative in one step and then worked from there to simplify. The quotient rule would have taken many more steps and left much more chance for errors. This useful technique is called logarithmic differentiation.