Powers and Logarithms

$\frac{d}{dx}(\ln x)=\frac{1}{x}$​
We want to prove the derivative of ln(x), so let’s write it as an equation and say that y=ln(x). Using the definition of a logarithm, we can rewrite this as:

$e^y=x$

Now we can implicitly differentiate both sides.

$\frac{d}{dx}(e^y=x)$

$\frac{d}{dx}{e}^y=\frac{d}{dx}x$

$e^y\frac{dy}{dx}=1$

Now, what was e^y again? We just said it was x! Let’s substitute that back in and we find that $\frac{dy}{dx}=\frac{1}{x}$ This is the derivative of ln(x) because we had set that equal to y. Imagine having to prove that using limits! No thanks.

$y=a^x$​
Before we take this derivative, let’s prep the equation a little bit by taking the natural log of both sides. You may have to recall your log rules for this.

$\ln y=x\cdot \ln a$

Now we can take the derivative of both sides​
$\frac{d}{dx}\ln y=\frac{d}{dx}x\cdot \ln a$​
$\frac{dy}{dx}\cdot \frac{1}{y}=1\cdot \ln a$​
Remember that ln(a) is just a constant! Simplifying this, multiplying by y on both sides, and substituting y= a^x gives us:

$\frac{\mathit{d}}{\mathit{d}\mathit{x}}\mathbf{(}{\mathit{a}}^{\mathit{x}}\mathbf{)}\mathbf{=}{\mathit{a}}^{\mathit{x}}\mathbf{\cdot }\mathbf{ln}\mathbf{(}\mathit{a}\mathbf{)}$

$y={\log }_{a}x$​
$a^y=x$​
Now we can take the derivative of both sides

$\frac{d}{dx}{a}^y=\frac{d}{dx}x$​
$\frac{dy}{dx}\cdot a^y\cdot \ln a=1$​
Remembering that since a^y= x, we can isolate $\frac{dy}{dx}$ and get

$\frac{dy}{dx}=\frac{1}{x\cdot \ln a}$

Let’s multiply both sides by y first so that we won’t have to take the quotient rule:

$y^2=x\cdot {\log }_{2}x+3^{\sin x}$​
Now we can take the derivative of both sides​
$\frac{d}{dx}\left(y^2=x\cdot {\log }_{2}x+3^{\sin x}\right)$​
$2y\cdot \frac{dy}{dx}=1\cdot {\log }_2(x)+x\cdot \frac{1}{x\cdot \ln 2}+3^{\sin x}\cdot \ln 3\cdot \cos x$​
Since we are getting into complex derivatives, let’s create a key for each rule.

$\text{Magenta}$ - Quotient Rule​
$\text{Green}$ - Chain Rule​
$\text{Blue}$ - Product Rule

$\frac{dy}{dx}=\frac{{\log }_{2}x+\cancel{x}\cdot \frac{1}{\cancel{x}\cdot \ln 2}+3^{\sin x}\cdot \ln 3\cdot \cos x}{2y}$

We need to figure out what y is equal to. Since​
$y^2=x{\log }_{2}x+3^{\sin x}$,

$y=\sqrt{x{\log }_2(x)+3^{\sin x}}$​
For simplicity’s sake, we’ll only be concerned with the positive square root. Substituting this equation into the derivative gives:

$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\mathbf{=}\frac{{\mathbf{log}}_{\mathbf{2}}\mathit{x}\mathbf{+}\frac{\mathbf{1}}{\mathbf{ln}\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{sin}\mathit{x}}\mathbf{\cdot }\mathbf{ln}\mathbf{3}\mathbf{\cdot }\mathbf{cos}\mathit{x}}{\mathbf{2}\sqrt{\mathit{x}\mathbf{\cdot }{\mathbf{log}}_{\mathbf{2}}\mathbf{(}\mathit{x}\mathbf{)}\mathbf{+}{\mathbf{3}}^{\mathbf{sin}\mathit{x}}}}$