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Practice Problems for 3/11

  1. Let \(f(x)=5x^2-10x+4\), for \(x\geq 1\). Find the value of the derivative of \(f^{-1}(x)\) at \(x=4=f(2)\).
    2. First, check that the requirements of Theorem 3.8.3 are satisfied. \(f(x)\) is differentiable and \(f'(x)=10x-10\). \(f'(1)=0\) but since we are interested in the derivative of \(f^{-1}\) at \(f(2)\), this is not a problem. (i.e. \(f'(2)\neq 0\), so the conditions of the theorem are satisfied at the point we're interested in.) Now, we can apply Theorem 3.3.8. \[ \frac{d}{dx}f^{-1}(x)|_{x=4}=\frac{1}{f'(f^{-1}(x))}|_{x=4}=\frac{1}{f'(f^{-1}(4))}=\frac{1}{f'(2)}=\frac{1}{10}. \]
  2. Find the derivative of the following with respect to the independent variable.

    1. \(f(x)=\left(\frac{3}{4}\right)^{2x}\)
      2. This is a composite function. The outside function is \( (3/4)^x\) and the inside function is \(2x\). Applying the chain rule gives, \[ \frac{d}{dx}f(x)=(3/4)^{2x}\ln(3/4)\frac{d}{dx}2x=2\ln(3/4)(3/4)^{2x}. \]
    2. \(g(s)=\log_5(s^2+s),\quad s>0\)
      3. This is a composite function. The outside function is \(\log_5(s)\) and the inside function is \(s^2+s\). By the chain rule, \[ \frac{d}{ds}g(s)=\frac{1}{\ln(5)(s^2+s)}\frac{d}{ds}s^2+s=\frac{2s+1}{\ln(5)(s^2+s)} \]
    3. \(s(t)=\log_3(3^{t^4})\)
      4. Since \(\log_3(t)\) and \(3^t\) are inverses, \(s(t)=t^4\). So, \(\frac{d}{dt}s(t)=4t^3.\)
  3. Use the product rule to find the derivative of \(y=x^2(7x^3+4)^6(x^2+1)\). (Or at least carry our the first two or three steps to get an idea of how using the product rule within the product rule would go.)
    4. First, write \(y=f(x)g(x)\) with \(f(x)=x^2\) and \(g(x)=(7x^3+4)^6(x^2+1)\). Then by the product rule, \[ \frac{dy}{dx}=2x(7x^3+4)^6(x^2+1)+x^2\frac{d}{dx}(7x^3+4)^6(x^2+1). \] Since \(g(x)=h(x)k(x)\) with \(h(x)=(7x^3+4)^6\) and \(k(x)=x^2+1\), we must use the product rule to find the derivative of \(g(x)\). Also, since \(h(x)\) is a composite function, we'll need to use the chain rule inside of the product rule.
    \(\begin{align*} \frac{d}{dx}(7x^3+4)^6(x^2+1)&=6(7x^3+4)^5(\frac{d}{dx}7x^3+4)(x^2+1)+(7x^3+4)^6(2x)\\ &=126x^2(7x^3+4)^5(x^2+1)+2x(7x^3+4)^6. \end{align*}\)
    Plugging this in for \(g'(x)\) implies
    \(\begin{align*} \frac{dy}{dx}&=2x(7x^3+4)^6(x^2+1)+x^2\frac{d}{dx}(7x^3+4)^6(x^2+1)\\ =&2x(7x^3+4)^6(x^2+1)+126x^4(7x^3+4)^5(x^2+1)+2x^3(7x^3+4)^6. \end{align*}\)