Exercise Set 10 Answers

  1. Find \(\dfrac{dy}{dx}\) for the following by implicit differentiation.

    1. \(x^2y+3xy^3-x=3\)
    2. \(\sin(x^2y^2)=x\)
    3. \(xy^2=y+e^{xy}\)

    Answers:

    1. \[\begin{align*} 2xy+x^2\frac{dy}{dx}+3y^3+9xy^2\frac{dy}{dx}-1&=0\\ (x^2+9xy^2)\frac{dy}{dx}&=1-2xy-3y^3\\ \frac{dy}{dx}&=\frac{1-2xy-3y^3}{x^2+9xy^2} \end{align*}\]

    2. \[\begin{align*} \cos(x^2y^2)\left(2xy^2+2x^2y\frac{dy}{dx}\right)&=1\\ \frac{dy}{dx}&=\frac{1-2xy^2\cos(x^2y^2)}{2x^2y\cos(x^2y^2)} \end{align*}\]

    3. \[\begin{align*} \frac{d}{dx}(xy^2)&=\frac{d}{dx}(y+e^{xy})\\ 2xy\frac{dy}{dx}+y^2&=\frac{dy}{dx}+\frac{d}{dx}e^{xy}\\ 2xy\frac{dy}{dx}+y^2&=\frac{dy}{dx}+e^{xy}\frac{d}{dx}(xy)\\ 2xy\frac{dy}{dx}+y^2&=\frac{dy}{dx}+e^{xy}\left(x\frac{dy}{dx}+y\right)\\ 2xy\frac{dy}{dx}-xe^{xy}\frac{dy}{dx}-\frac{dy}{dx}&=ye^{xy}-y^2\\ \frac{dy}{dx}(2xy-xe^{xy}-1)&=ye^{xy}-y^2\\ \frac{dy}{dx}&=\frac{ye^{xy}-y^2}{(2xy-1-xe^{xy})}. \end{align*}\]

  2. Consider the curve \(x^2+3xy+y^2+2x-7=0\). Use implicit differentiation to find \(\dfrac{dy}{dx}\) and hence compute the equation of the tangent at the point \((1,1)\).

    Answers:

    \[\begin{align*} 2x+3y+3x\frac{dy}{dx}+2\frac{dy}{dx}+2&=0\\ \frac{dy}{dx}&=-\frac{2x+3y+2}{3x+2}.\\ \end{align*}\] At the point \((1,1)\) \[\begin{equation*} \frac{dy}{dx}\Bigr|_{\substack{(1,1)}}=-\frac{2+3+2}{3+2}=-\frac{7}{5} \end{equation*}\] then the equation of the tangent line is \[\begin{equation*} y=1-\frac{7}{5}(x-1). \end{equation*}\]

  3. Find the slope of the circle \(x^2+y^2=25\) at the point \((-3,-4)\). Hence find the equations of the tangent and normal at that point.

    Answers: By implicit differentiation \[2x+2y\frac{dy}{dx}=0\] hence \[\frac{dy}{dx}=-\frac{x}{y}\] and at \((-3,-4)\) \[\frac{dy}{dx}=-\frac{3}{4}.\]

    The equation of the tangent line is given by \[y-(-4)=-\frac{3}{4}(x-(-3))\] or \[y=-\frac{3}{4}x-\frac{25}{4}.\]

    The equation of the normal line is \[y-(-4)=\frac{4}{3}(x-(-3))\] or \[y=\frac{4}{3}x.\]

  4. Find \(\dfrac{dy}{dx}\) and \(\dfrac{d^2y}{dx^2}\) for the following by implicit differentiation.

    1. \(y+\sin(y)=x\)
    2. \(x^2-xy+y^2=3\)

    Answers:

    1. \[\begin{align*} \frac{dy}{dx}+\cos(y)\frac{dy}{dx}&=1\\ \frac{dy}{dx}&=\frac{1}{1+\cos(y)} \end{align*}\]
    2. \[\begin{align*} 2x-y-x\frac{dy}{dx}+2y\frac{dy}{dx}&=0\\ (x-2y)\frac{dy}{dx}&=2x-y\\ \frac{dy}{dx}&=\frac{2x-y}{x-2y} \end{align*}\]

  5. Find \(\dfrac{dy}{dx}\) and \(\dfrac{d^2y}{dx^2}\) for the following parametric equations.

    1. \(y=\cos(2t)\) and \(x=\sin(t)\)
    2. \(y=\dfrac{3+2t}{1+t}\) and \(x=\dfrac{2-3t}{1+t}\)
    3. \(y=3\sin(\theta)-\sin^3(\theta)\) and \(x=\cos^3(\theta)\)

    Answers:

    1. \[\begin{align*} \frac{dy}{dt}=-2\sin(2t)&&\frac{dx}{dt}=\cos(t) \end{align*}\] The first derivative is \[\begin{align*} \frac{dy}{dx}&=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\\ &=\frac{-2\sin(2t)}{\cos(t)}\\ &=\frac{-4\sin(t)\cos(t)}{\cos(t)}\\ &=-4\sin(t). \end{align*}\] The second derivative is (using the chain rule) \[\begin{align*} \frac{d^2y}{dx^2}&=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}\\ &=\frac{d}{dt}(-4\sin(t))\cdot \frac{1}{\cos(t)}\\ &=-4\cos(t)\cdot \frac{1}{\cos(t)}\\ &=-4. \end{align*}\]

    2. \[\begin{align*} \frac{dy}{dt}=-\frac{1}{(1+t)^2}&&\frac{dx}{dt}=-\frac{5}{(1+t)^2} \end{align*}\] The first derivative is \[\begin{align*} \frac{dy}{dx}&=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\\ &=\frac{1}{(1+t)^2}\cdot \frac{(1+t)^2}{5}\\ =\frac{1}{5} \end{align*}\] then the second derivative \[\begin{equation*} \frac{d^2y}{dx^2}=0. \end{equation*}\]

    3. \[\begin{align*} \frac{dy}{d\theta}=3\cos^3(\theta)&&\frac{dx}{d\theta}=-3\cos^2(\theta)\sin(\theta) \end{align*}\] The first derivative is \[\begin{align*} \frac{dy}{dx}&=\frac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}\\ &=\frac{3\cos^3(\theta)}{-3\cos^2(\theta)\sin(\theta)}\\ &=-\cot(t) \end{align*}\] and the second derivative \[\begin{align*} \frac{d^2y}{dx^2}&=\frac{d}{d\theta}\left(\frac{dy}{dx}\right)\frac{d\theta}{dx}\\ &=\frac{d}{d\theta}(-\cot(t))\cdot \frac{1}{-3\cos^2(\theta)\sin(\theta)}\\ &=\frac{1}{\sin^2(\theta)}\cdot \frac{1}{-3\cos^2(\theta)\sin(\theta)}\\ &=\frac{-1}{3\cos^2(\theta)\sin^3(\theta)}, \end{align*}\] where we have used \[\begin{align*} \frac{d}{d\theta}(-\cot(t))&=\frac{d}{d\theta}\frac{-\cos(\theta)}{\sin(\theta)}\\ &=\frac{\sin(\theta)\sin(\theta)-(-\cos(\theta))\cos(\theta)}{\sin^2(\theta)}\\ &=\frac{1}{\sin^2(\theta)}. \end{align*}\]

  6. Find the derivative of the following inverse functions by implicit differentiation.

    1. \(y=\sin^{-1}(x)\)
    2. \(y=\cosh^{-1}(3x)\)
    3. \(y=\tan^{-1}(4x^2)\)

    Answers:

    1. We have that \[\begin{equation*} \sin(y)=x \end{equation*}\] then by implicit differentiation \[\begin{align*} \cos(y)\dfrac{dy}{dx}=1\\ \dfrac{dy}{dx}=\dfrac{1}{\cos(y)}. \end{align*}\] Using the identity \(\cos^2(y)+\sin^2(y)=1\) \[\begin{align*} \cos(y)&=\sqrt{1-\sin^2(y)}\\ &=\sqrt{1-x^2} \end{align*}\] so, \[\begin{equation*} \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}. \end{equation*}\]

    2. We have \[\begin{equation*} \cosh(y)=3x \end{equation*}\] then by implicit differentiation \[\begin{align*} \sinh(y)\dfrac{dy}{dx}&=3\\ \dfrac{dy}{dx}&=\dfrac{3}{\sinh(y)}. \end{align*}\]

      Since \(\cosh^2(y)-\sinh^2(y)=1\) \[\begin{equation*} \sinh(y)=\sqrt{\cosh^2(y)-1}=\sqrt{(3x)^2-1}. \end{equation*}\] So, \[\begin{equation*} \dfrac{dy}{dx}=\frac{3}{\sqrt{9x^2-1}}. \end{equation*}\]

    3. We have \[\begin{equation*} \tan(y)=4x^2. \end{equation*}\] By implicit differentiation, \[\begin{align*} \sec^2(y)\dfrac{dy}{dx}&=8x\\ \dfrac{dy}{dx}&=\dfrac{8x}{\sec^2(y)} \end{align*}\] Using \[\begin{align*} \cos^2(y)+\sin^2(y)&=1\\ 1+\tan^2(y)&=\sec^2(y) \end{align*}\] we have \(\sec^2(y)=1+(4x^2)^2\), and then \[\begin{equation*} \dfrac{dy}{dx}=\dfrac{8x}{1+16x^4}. \end{equation*}\]

  7. Find \(\dfrac{dy}{dx}\) for the following using logarithmic differentiation.

    1. \(y=a^x\)
    2. \(y=\dfrac{(x-4)^7(2x+3)^2}{(4x+7)^3}\)

    Answers:

    1. Taking logs and differentiating implicitly \[\begin{align*} \ln(y)=\ln(a^x)&=x\ln(a)\\ \frac{1}{y}\dfrac{dy}{dx}&=\ln(a)\\ \dfrac{dy}{dx}&=y\ln(a)\\ \dfrac{dy}{dx}&=a^x\ln(a). \end{align*}\]

    2. Taking logs \[\begin{align*} \ln(y)&=\ln\left[\dfrac{(x-4)^7(2x+3)^2}{(4x+7)^3}\right]\\ &=\ln(x-4)^7+\ln(2x+3)^2-\ln(4x+7)^3\\ &=7\ln(x-4)+2\ln(2x+3)-3\ln(4x+7) \end{align*}\] Then differentiating \[\begin{align*} \frac{1}{y}\dfrac{dy}{dx}&=\frac{7}{x-4}+\frac{2}{2x+3}\cdot 2-\frac{3}{4x-7}\cdot 4\\ \dfrac{dy}{dx}&=\dfrac{(x-4)^7(2x+3)^2}{(4x+7)^3}\left[ \frac{7}{x-4}+\frac{4}{2x+3}-\frac{12}{4x-7}\right] \end{align*}\]

  8. Find \(\dfrac{dy}{dx}\) for the following.

    1. \(y=x^x\)
    2. \(y=(\tanh(x))^x\)
    3. \(x^3+\sin(xy)=xy^2\)

    Answers:

    1. First take the natural logarithm: \[\begin{equation*} \ln(y)=x\ln(x), \end{equation*}\] then differentiate implicitly: \[\begin{align*} \frac{d}{dx}(\ln(y))&=\frac{d}{dx}(x\ln(x))\\ \frac{1}{y}\frac{dy}{dx}&=\frac{dx}{dx}\ln(x)+x\frac{d}{dx}\ln(x)\\ \frac{1}{y}\frac{dy}{dx}&=\ln(x)+x\frac{1}{x}\\ \frac{dy}{dx}&=y(\ln(x)+1)=x^x(\ln(x)+1). \end{align*}\]

      Important! The derivative of \(x^x\) is not \(x\cdot x^{x-1}=x^x\). We can’t treat variable exponents like constant exponents!

    2. Taking logs \[\begin{equation*} \ln(y)=x\ln(\tanh(x)), \end{equation*}\] then differentiating \[\begin{align*} \frac{1}{y}\dfrac{dy}{dx}&=\ln(\tanh(x))+\frac{x}{\tanh(x)}\textrm{sech}^2(x)\\ &=\ln(\tanh(x))+\frac{x\cosh(x)}{\sinh(x)\cosh^2(x)}\\ &=\ln(\tanh(x))+\frac{x}{\sinh(x)\cosh(x)}\\ &=\ln(\tanh(x))+\frac{2x}{\sinh(2x)}.\\ \end{align*}\] So, \[\begin{equation*} \dfrac{dy}{dx}=(\tanh(x))^x\left[ \ln(\tanh(x))+\frac{2x}{\sinh(2x)}\right]. \end{equation*}\]

    3. By implicit differentiation and using the product rule \[\begin{align*} 3x^2+\cos(xy)\left(y+x\dfrac{dy}{dx} \right)&=y^2+2xy\dfrac{dy}{dx}\\ 3x^2+y\cos(xy)-y^2&=2xy\dfrac{dy}{dx}-x\cos(xy)\dfrac{dy}{dx} \end{align*}\] so, \[\begin{equation*} \dfrac{dy}{dx}=\frac{3x^2+y\cos(xy)-y^2}{2xy-x\cos(xy)} \end{equation*}\]