Derivatives of Logarithmic and Exponential Functions

This lesson works through five proof-style problems on the derivatives of logarithmic and exponential functions. Each one asks us to differentiate a given expression and show that it equals a required result, leaning on the chain rule, the quotient rule, and the laws of logarithms.

Logarithm of a square-root expression

We prove that

$\frac{d}{dx}\,\log\!\left(x + \sqrt{x^2 - a^2}\right) = \frac{1}{\sqrt{x^2 - a^2}}.$

This is a function of a function, so we apply the chain rule. Since $\frac{d}{dx}\log u = \frac{1}{u}\cdot \frac{du}{dx}$,

$\frac{d}{dx}\,\log\!\left(x + \sqrt{x^2 - a^2}\right) = \frac{1}{x + \sqrt{x^2 - a^2}}\cdot \frac{d}{dx}\!\left(x + \sqrt{x^2 - a^2}\right).$

The inner derivative. Here $\frac{d}{dx}\sqrt{x^2 - a^2} = \frac{1}{2\sqrt{x^2 - a^2}}\cdot 2x = \frac{x}{\sqrt{x^2 - a^2}}$, so

$\frac{d}{dx}\!\left(x + \sqrt{x^2 - a^2}\right) = 1 + \frac{x}{\sqrt{x^2 - a^2}} = \frac{\sqrt{x^2 - a^2} + x}{\sqrt{x^2 - a^2}}.$

Putting it together. The factor $x + \sqrt{x^2 - a^2}$ in the denominator cancels with the matching numerator:

$\frac{1}{x + \sqrt{x^2 - a^2}}\cdot \frac{x + \sqrt{x^2 - a^2}}{\sqrt{x^2 - a^2}} = \frac{1}{\sqrt{x^2 - a^2}}.$

A log simplified with the conjugate

We prove that

$\frac{d}{dx}\,\log\!\left(\frac{\sqrt{x+1} + \sqrt{x-1}}{\sqrt{x+1} - \sqrt{x-1}}\right) = \frac{1}{\sqrt{x^2 - 1}}.$

Rationalise first. Multiply the numerator and denominator inside the log by the conjugate of the denominator, $\sqrt{x+1} + \sqrt{x-1}$:

$\frac{\left(\sqrt{x+1} + \sqrt{x-1}\right)^2}{\left(\sqrt{x+1}\right)^2 - \left(\sqrt{x-1}\right)^2}.$

The denominator is $(x+1) - (x-1) = 2$. The numerator expands to

$(x+1) + (x-1) + 2\sqrt{(x+1)(x-1)} = 2x + 2\sqrt{x^2 - 1}.$

So the fraction becomes

$\frac{2x + 2\sqrt{x^2 - 1}}{2} = x + \sqrt{x^2 - 1}.$

Now differentiate. The problem reduces to

$\frac{d}{dx}\,\log\!\left(x + \sqrt{x^2 - 1}\right),$

which is the same chain-rule pattern as before with $a = 1$:

$= \frac{1}{x + \sqrt{x^2 - 1}}\cdot \frac{x + \sqrt{x^2 - 1}}{\sqrt{x^2 - 1}} = \frac{1}{\sqrt{x^2 - 1}}.$

An exponential rewritten with the laws of logarithms

If $y = e^{3\log x + 2x}$, prove that $\dfrac{dy}{dx} = x^2 e^{2x}(2x + 3)$.

Simplify first. Using $e^{m+n} = e^m\, e^n$ and $m\log x = \log x^m$,

$y = e^{\log x^3}\, e^{2x}.$

Since $e^{\log m} = m$, this gives $y = x^3 e^{2x}$.

Product rule.

$\frac{dy}{dx} = x^3\cdot e^{2x}\cdot 2 + e^{2x}\cdot 3x^2 = x^2 e^{2x}(2x + 3).$

Nine to a base-three log

If $y = 9^{\log_3 x}$, prove that $\dfrac{dy}{dx} = 2x$.

Write $9 = 3^2$, so

$y = 3^{2\log_3 x} = 3^{\log_3 x^2}.$

Using the law $a^{\log_a m} = m$, this collapses to $y = x^2$. Differentiating,

$\frac{dy}{dx} = 2x.$

A quotient of exponentials

If $y = \dfrac{e^x + e^{-x}}{e^x - e^{-x}}$, prove that $\dfrac{dy}{dx} = 1 - y^2$.

Quotient rule. With numerator $e^x + e^{-x}$ and denominator $e^x - e^{-x}$,

$\frac{dy}{dx} = \frac{\left(e^x - e^{-x}\right)\left(e^x - e^{-x}\right) - \left(e^x + e^{-x}\right)\left(e^x + e^{-x}\right)}{\left(e^x - e^{-x}\right)^2}.$

Split the fraction. This is

$\frac{\left(e^x - e^{-x}\right)^2}{\left(e^x - e^{-x}\right)^2} - \frac{\left(e^x + e^{-x}\right)^2}{\left(e^x - e^{-x}\right)^2} = 1 - \left(\frac{e^x + e^{-x}}{e^x - e^{-x}}\right)^2.$

The second term is exactly $y^2$, so

$\frac{dy}{dx} = 1 - y^2.$

Key takeaways

• To differentiate $\log\!\left(x + \sqrt{x^2 - a^2}\right)$, apply the chain rule; the awkward factor cancels and leaves $\frac{1}{\sqrt{x^2 - a^2}}$.
• Rationalising a fraction with its conjugate can turn a complicated logarithm into a simple one before you differentiate.
• The laws of logarithms ($e^{\log m} = m$ and $a^{\log_a m} = m$) reduce many exponential expressions to plain powers of $x$.
• The quotient rule on $\frac{e^x + e^{-x}}{e^x - e^{-x}}$ produces a result you can read back in terms of $y$, namely $1 - y^2$.