Exercises - Derivatives Involving Inverse Trigonometric Functions and Logarithms

  1. Suppose $k$ is a nonzero real number. Find $\displaystyle{\frac{d}{dx} \ln (kx)}$ and compare it to the derivative of $\ln x$. Are you surprised? Find a second explanation for what you see.

    By the chain rule, $\displaystyle{\frac{d}{dx} \ln (kx) = \frac{1}{kx} \cdot {k} = \frac{1}{x}}$.

    This of course is identical to the derivative of $\ln x$.

    Note that by the properties of logarithms, $\ln (kx) = \ln x + \ln k$. So taking the derivative of $\ln (kx)$ is the same as taking the derivative of $\ln x$ plus a constant -- and the constant vanishes under differentiation!

  2. Differentiate $\displaystyle{y = \frac{x^3 \sqrt[3]{x^2 + 1}}{(3x+2)^5}}$.

    We use logarithmic differentiation to simplify things...

    $$\begin{array}{rcl} y &=& \displaystyle{\frac{x^3 \sqrt[3]{x^2 + 1}}{(3x+2)^5}}\\\\ &=& \displaystyle{{\large e}^{\displaystyle{\ln \left[ \frac{x^3 \sqrt[3]{x^2 + 1}}{(3x+2)^5} \right]}}}\\\\ &=& \displaystyle{{\large e}^{\displaystyle{\left[ 3\ln x + \frac{1}{3} \ln (x^2+1) - 5\ln (3x+2) \right]}}}\\\\ \end{array}$$

    Now we differentiate this last expression using the chain rule to find

    $$y' = \displaystyle{{\large e}^{\displaystyle{\left[ 3\ln x + \frac{1}{3} \ln (x^2+1) - 5\ln (3x+2) \right]}} \cdot \left( \frac{3}{x} + \frac{2x}{3(x^2+1)} - \frac{15}{3x+2} \right)}$$

    Finally, upon realizing the power of $e$ above is identical to the original expression for $y$, we have

    $$y' = \frac{x^3 \sqrt[3]{x^2 + 1}}{(3x+2)^5} \cdot \left( \frac{3}{x} + \frac{2x}{3(x^2+1)} - \frac{15}{3x+2} \right)$$
  3. Find $\displaystyle{\frac{dy}{dx}}$ given that $\sqrt{x} + \sqrt{y} = 1$.

    Differentiating both sides with respect to $x$, we have

    $$\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0$$

    Now solving for $\displaystyle{\frac{dy}{dx}}$, we find

    $$\frac{dy}{dx} = \frac{\displaystyle{-\frac{1}{2\sqrt{x}}}}{\displaystyle{\frac{1}{2\sqrt{y}}}} = -\frac{\sqrt{y}}{\sqrt{x}} = -\sqrt{\frac{y}{x}}$$
  4. Find the derivative of $\displaystyle{f(m) = \arcsin \left( \frac{2m}{m^2 + 1} \right)}$ and determine all points where $f(m)$ is differentiable.

    Using the derivative formula for arcsine, the chain rule, and the quotient rule, we get

    $$\frac{d}{dx} \arcsin \left( \frac{2m}{m^2 + 1} \right) = \frac{1}{\sqrt{\displaystyle{1 - \left( \frac{2m}{m^2+1} \right)^2}}} \cdot \frac{(m^2+1) \cdot 2 - 2m \cdot 2m}{(m^2+1)^2}$$

    With a bit of algebra, we transform the above into

    $$\frac{d}{dx} \arcsin \left( \frac{2m}{m^2 + 1} \right) = \frac{m^2+1}{|m^2-1|} \cdot \frac{-2(m^2-1)}{(m^2+1)^2} = -2 \cdot \frac{|m^2-1|}{m^4-1}$$

    Note, the last simplification takes advantage of the fact that $\displaystyle{\frac{-x}{|x|} = \frac{|x|}{-x}}$.

    Finally, considering the domain of $f'$ and where the absolute value changes behavior, we see that $f'(m)$ is undefined for $m = \pm 1$, but differentiable at all other values of $m$.