Derivative Inverse

Now that we know the chain rule and implicit differentiation, we can calculate the derivative of inverse functions. This is a pretty useful skill!

Example: finding the derivative of the inverse of $e^x$: The invese of $e^x$ is the natural logarithm $\ln(x)$. That means, if $y=\ln(x)$ then $e^y=x$. We can differentiate this equation to get an expression for $\frac{dy}{dx}$: $$e^yy^\prime =1$$ Solving for $y^\prime$ gives $y^\prime = 1/e^y$. But remember $e^y$ is just $x$! So this is $1/x$

$$\frac{d}{dx}\ln(x)=\frac{1}{x}$$

Another way to do this same problem is to more explicitly use the chain rule (which, remember is hiden inside of our method for implicit differentiation): Since $e^x$ and $\ln(x)$ are inverses, we have $e^(\ln(x))=x$ and $\ln(e^x)=x$ on the domains where they are defined. We can take the derivative of either of these expressions directly, and solve the result for the mystery derivative:

$$\frac{d}{dx}e^{\ln(x)}=e^{\ln(x)}\ln(x)^\prime = 1$$ Then recall that $e^{\ln(x)}=x$ and divide by it!

$$\frac{d}{dx}\ln(e^x)=\ln^\prime(e^x)e^x=1$$ Thus, the derivative of $\ln$ is something that when evaluated at $e^x$ gives $1/e^x$. When evaluated at $x$ then, it gives $1/x$!

The story of $\ln|x|$. The abovve arguments only show the derivative is 1/x for positive values of $x$. What about negative values? Can’t plug these into the logarithm. But - (graph it)! We see the derivative of $\ln|x|$ is $1/x$ for all nonzero $x$.

Derivative of the square root in this way:

We have already calculated the derivative of the square root in two different ways: directly from the definition, and then using the product rule. But we can also do it, using that the square root is the inverse of squaring! If $y=\sqrt{x}$ then $x=y^2$ and differentiating gives $1=2yy^\prime$ so $y^\prime=\frac{1}{2y}=\frac{1}{2\sqrt{x}}$.

Derivatives of inverse trigonometric functions

If $y=\arcsin(x)$ then $\sin(y)=x$ and differentiating we see $$\cos(y)y^\prime =1 $$ Now we just need to express $\cos(y)$ in terms of $x$: $$\cos y=\sqrt{1-\sin^2 y}=\sqrt{1-x^2}$$ Thus, $$\frac{d}{dx}\arcsin(x)=\frac{1}{\sqrt{1-x^2}$$

We can do $\arccos$ similarly: lets try arctangent. If $y=\arctan(x)$ then $x=\tan(y)$ and differentiating: $$1=\sec^2(y)y^\prime$$ Solving for $y^\prime$: $$y^\prime=\fra{1}{\sec^2(y)}=\frac{1}{1+\tan^2(y)}=\frac{1}{1+x^2}$$

Logarithmic Differentiation

Another use of implicit differentiation is called logarithmic differentiation: if an expression is too difficult to differentiate at the start, try taking the logarithm of everything involved and differentiating: this is helpful because logs turn exponentiation into multiplication, and multiplication into addition:

Example: take the derivative of $x^x$: $y=x^x$ looks impossible to differentiate with our known rules: we know what to do when the base is constant or when the exponent is constnat, but not what to do when both vary. So, take the logarithm: $\ln(y)=\ln(x^x)$ and simplify: $$\ln(y)=x\ln(x)$$ Take the derivative of this expression, and then solve for $y^\prime$!

$$\frac{1}{y}y^\prime = \ln(x)+1$$

$$y^\prime = x^x(1+\ln(x))$$

This trick lets us do many useful thigns, including (finally)! Prove the power rule for all real exponents! If $y=x^n$ then $\ln(y)=n\ln(x)$ and differentiating gives $\frac{1}{y}y^\prime = n\frac{1}{x}$$ Solving for $y^\prime$: $$y^\prime = n\frac{y}{x}=n\frac{x^n}{x}=nx^{n-1}$$

We can also use this to avoid crazy long applications of the product and quotient rules: $$y=\frac{x^{3/4}\sqrt{x^2+1}}{(3x+2)^5}$$ Taking the logarithm and simplifying: $$\ln(y)=\ln(x)^{3/4}+\ln(x^2+1)^{1/2}-\ln(3x+2)^5$$ $$\ln(y)=\frac{3}{4}\ln(x)+\frac{1}{2}\ln(x^2+1)-5\ln(3x+2)$$

Taking the derivative: $$\frac{y^\prime}{y}=\frac{3}{4}\frac{1}{x}+\frac{1}{2}\frac{2x}{x^2+1}-5\frac{3}{3x+2}$$

Then just multiply by $y$ (written as a function of $x$ it started as) and you’re done $$y^\prime = \frac{x^{3/4}\sqrt{x^2+1}}{(3x+2)^5}\left(\frac{3}{4x}+\frac{x}{x^2+1}-\frac{15}{3x+2}\right)$$