Back to Section Page Back to Problems Page

Practice Problems for 3/4

  1. Find \(\frac{dy}{dx}\) for \(y=e^{\sec x}\).

    2. Let \(u=g(x)=\sec x\) and \(y=f(u)=e^u\). Then \(\frac{dy}{du}=e^u\) and \(\frac{du}{dx}=\sec x \tan x\). By the chain rule, \[ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}=e^u\sec x \tan x=e^{\sec x}\sec x \tan x. \] OR
    y=\((f\circ g)(x)\), with \(f(x)=e^x\) and \(g(x)=\sec x\). \(f'(x)=e^x\), and \[ \frac{dy}{dx}=e^{\sec x} \frac{d}{dx}\sec x. \] Since \(\frac{d}{dx}\sec x=\sec x \tan x\), we have \[ \frac{dy}{dx}=e^{\sec x}\sec x\tan x. \]
  2. Each of the following functions can be written as \((f\circ g)(x)\). Define an appropriate \(f\) and \(g\) for each function.
    1. \(5^{-1/x}\)
      2. If \(f(x)=5^x\) and \(g(x)=-1/x\), then \(5^{-1/x}=(f\circ g)(x)\).
    2. \(\cot(x^3)\)
      3. \(f(x)=\cot x\) and \(g(x)=x^3\).
    3. \(\left( \frac{x-6}{x^7+9} \right)^4\)
      4. \(f(x)=x^4\) and \(g(x)=\frac{x-6}{x^7+9}\).

    If we wanted to take the derivative of any of the above functions, we would find \(f'(x)\) and \(g'(x)\) and apply the chain rule.

  3. Let \(y=\frac{x^2}{\sin x}\).
    1. Calculate \(\frac{dy}{dx}\) using the quotient rule.
      2. \[ \frac{dy}{dx}=\frac{(\frac{d}{dx}x^2)\sin x - x^2 \frac{d}{dx}\sin x}{(\sin x )^2}=\frac{2x\sin x -x^2\cos x}{\sin^2 x} \]
    2. Calculate \(\frac{dy}{dx}\) using the product rule by writing \(y=x^2(\sin x)^{-1}\).
      Since \((\sin x)^{-1}\) is a composition, we'll also need to apply the chain rule. \begin{align*} \frac{dy}{dx}&=(\frac{d}{dx} x^2)\sin x+x^2 \frac{d}{dx}(\sin x)^{-1}=2x (\sin x)^{-1} + x^2(-1)(\sin x)^{-2}\cos x\\ &=\frac{2x}{\sin x}-\frac{x^2\cos x}{\sin^2 x}=\frac{2x\sin x -x^2\cos x}{\sin^2 x}. \end{align*}

    You can always use the product and chain rule instead of the quotient rule. In a way it's easier because you can just remember the product rule and not the quotient rule.
  4. Let \(y=m(x)=\cos(\sqrt{x^4+x})\).

    1. \(m(x)\) is the composition of three functions. Find \(f,\;g,\;\) and \(h\) such that \(m(x)=(h\circ f\circ g)(x)\).
      2. \[ g(x)=x^4+x,\quad f(x)=\sqrt{x},\quad \textrm{and}\quad h(x)=\cos x. \]
    2. Find \(\frac{dy}{dx}\).
      3. Apply the chain rule: \[ \frac{dy}{dx}=-\sin(\sqrt{x^4+x})\frac{d}{dx}\sqrt{x^4+x}. \] Since \(\sqrt{x^4+x}\) is also a composition, the chain rule must be applied again. \[ \frac{d}{dx} \sqrt{x^4+x}=\frac{1}{2\sqrt{x^4+x}}\frac{d}{dx}x^4+x=\frac{4x^3+1}{2\sqrt{x^4+x}}. \] Plugging this into the above expression for the derivative of \(y\) gives, \[ \frac{dy}{dx}=-\sin(\sqrt{x^4+x})\frac{4x^3+1}{2\sqrt{x^4+x}}. \]