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Practice Problems for 3/6

  1. Use implicit differentiation to find \(\frac{dy}{dx}\) for each of the following equations.
    1. \(xy=7\)
      2. Derivative of LHS: Use the product rule. \[ \frac{d}{dx}xy= (1)y+x\frac{d}{dx}y=y+x(1)\frac{dy}{dx} \] Derivative of RHS: \[ \frac{d}{dx}7=0 \] Solve for \(\frac{dy}{dx}\): \[ y+x\frac{dy}{dx}=0\implies \frac{dy}{dx}=\frac{-y}{x} \]
    2. \(\sin(x+y)=y^2\cos x\)
      3. Derivative of LHS: Chain rule \[ \frac{d}{dx}\sin(x+y)=\cos(x+y)\frac{d}{dx}(x+y)=\cos(x+y)(1+\frac{dy}{dx}) \] Derivative of RHS: Product rule \[ \frac{d}{dx}y^2\cos x=2y\frac{dy}{dx}+y^2(-\sin x)=2y\frac{dy}{dx}-y^2\sin x \] Solve for \(\frac{dy}{dx}\):
      \( \begin{align*} \cos(x+y)(1+\frac{dy}{dx}) &= 2y\frac{dy}{dx}-y^2\sin x \\ &\implies \frac{dy}{dx}(2y\cos x-\cos(x+y))=\cos(x+y)-y^2\sin x\\ &\implies \frac{dy}{dx}=\frac{\cos(x+y)-y^2\sin x}{2y\cos x-\cos(x+y)} \end{align*} \)
    3. \(e^{xy}=e^{4x}-e^y\)
      4. Derivative of LHS: Chain Rule \[ \frac{d}{dx}e^{xy}=e^{xy}\frac{d}{dx}xy=e^{xy}(y+x\frac{dy}{dx}) \] Derivative of RHS: \[ \frac{d}{dx}e^{4x}-\frac{d}{dx}e^y=4e^{4x}-e^y\frac{dy}{dx} \] Solve for \(\frac{dy}{dx}\):
      \( \begin{align*} e^{xy}(y+x\frac{dy}{dx})&=4e^{4x}-e^y\frac{dy}{dx}\\ &\implies \frac{dy}{dx}(xe^{xy}+e^y)=4e^{4x}-ye^{xy}\\ &\implies \frac{dy}{dx}=\frac{4e^{4x}-ye^{xy}}{xe^{xy}+e^y} \end{align*} \)

  2. The folium of Descarte (pictured below) is given by the equation \(x^3+y^3=6xy\)
    1. Find the equation for the tangent line to the folium at the point (3,3).
      2. Derivative of LHS: \[ \frac{d}{dx}x^3+y^3=3x^2+3y^2\frac{dy}{dx} \] Derivative of RHS: Product rule \[ \frac{d}{dx}6xy=6y+6x\frac{dy}{dx} \] Solve for \(\frac{dy}{dx}\): \[ 3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx} \implies \frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x} \]
    2. Find the equation for the normal line to the folium at the point (3,3)
      3. The slope of the tangent line at (3,3) is \[ m=\frac{6(x)-3(3^2)}{x(3^2)-6(3)}=\frac{-9}{9}=-1. \] This implies the slope of the normal line is \(-1/m=1\). The line with slope 1 through (3,3) is given by the equation \(y=x.\)