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Practice Problems for 3/13

  1. Use logarithmic differentiation find the derivative of \(y=x^2(7x^3+4)^6(x^2+1)\).
    Compare with your attempt to do this with the product rule.

    2. First check that the function we want to plug into the natural log is actually postive. (Here it is.) So \(\ln y=\ln[x^2(7x^3+4)^6(x^2+1)]=2\ln x+6\ln(7x^3+4)+\ln(x^2+1)\). Use implicit differentiation to find the derivative of y.
    1. \[ \frac{d}{dx}\ln y=\frac{1}{y}\frac{dy}{dx} \] 2.
    \( \begin{align*} \frac{d}{dx}\ln[x^2(7x^3+4)^6(x^2+1)]&=\frac{d}{dx}2\ln x+\frac{d}{dx}6\ln(7x^3+4)+\frac{d}{dx}\ln(x^2+1)\\ &= \frac{2}{x}+\frac{6(21x^2)}{7x^3+4}+\frac{2x}{x^2+1} \end{align*} \)
    3. Solve for \(\frac{dy}{dx}\).
    \( \begin{align*} \frac{dy}{dx}&=y\left(\frac{2}{x}+\frac{126x^2}{7x^3+4}+\frac{2x}{x^2+1}\right)\\ &=x^2(7x^3+4)^6(x^2+1)\left(\frac{2}{x}+\frac{126x^2}{7x^3+4}+\frac{2x}{x^2+1}\right) \end{align*}\)
  2. Find \(\frac{dy}{dx}\) for \(y=\sin(x)^{\sqrt{x}}\), \(x\in (0,\pi)\).

    3. Since \(y>0\) we can use logarithmic differentiation, which is good because we don't have another tool to find the derivative of this type of function. \[ \ln y= \ln (\sin x ^{\sqrt{x}})=\sqrt{x}\ln (\sin x) \] Differentiate both sides of the equation with respect to \(x\).
    1. \[ \frac{d}{dx}\ln y=\frac{1}{y}\frac{dy}{dx} \] 2. \[ \frac{d}{dx}\sqrt{x}\ln (\sin x)=\frac{1}{2\sqrt{x}}\ln(\sin x)+\sqrt{x}\frac{1}{\sin x}\frac{d}{dx}\sin x=\frac{1}{2\sqrt{x}}\ln(\sin x)+\sqrt{x}\frac{1}{\sin x}\cos x \] 3. \[ \frac{dy}{dx}=y\left(\frac{1}{2\sqrt{x}}\ln(\sin x)+\sqrt{x}\frac{1}{\sin x}\cos x\right)=\sin x^{\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\ln(\sin x)+\sqrt{x}\frac{1}{\sin x}\cos x\right). \]
  3. Find \(\frac{dy}{dx}\) for \(y=\frac{\csc(x)(x^5+7x^{-2})}{x^4+x+6}\), \(x\in(0,\pi)\).

    4. We could use the quotient rule, but since there is a product in the numerator, lets try logarithmic differentiation. (You can of course do it both ways to compare!) \[ \ln y= \ln \left( \frac{\csc(x)(x^5+7x^{-2})}{x^4+x+6}\right)=\ln (\csc(x)(x^5+7x^{-2}))-\ln(x^4+x+6) \] Differentiate both sides of the equation with respect to \(x\).
    1. \[ \frac{d}{dx}\ln y=\frac{1}{y}\frac{dy}{dx} \] 2.
    \(\begin{align*} \frac{d}{dx}\ln (\csc(x)(x^5+7x^{-2}))-\ln(x^4+x+6)& =\frac{d}{dx}\ln (\csc x)+\ln(x^5+7x^{-2})-\ln(x^4+x+6)\\ &=\frac{-\csc x \cot x}{\csc x}+\frac{5x^4-14x^{-3}}{x^5+7x^{-2}}-\frac{4x^3+1}{x^4+x+6}\\ &=-\cot x+\frac{5x^4-14x^{-3}}{x^5+7x^{-2}}-\frac{4x^3+1}{x^4+x+6} \end{align*}\)
    3. Solve for \(\frac{dy}{dx}\). \[ \frac{dy}{dx}=\frac{\csc(x)(x^5+7x^{-2})}{x^4+x+6} \left(-\cot x+\frac{5x^4-14x^{-3}}{x^5+7x^{-2}}-\frac{4x^3+1}{x^4+x+6} \right) \]