Pick any exercise in Chapter 12 of the calculus textbook. Post the solution or your attempt. Discuss any issues you might have had.
The problems I will be picking up are: Chapter #12, Exercises #31, page #679.
\(f(x, y, z) = x^2 e^{2y−3z}\)
Find
Find \(f_x\), \(f_y\), \(f_z\), \(f_{yz}\) and \(f_{zy}\).
\(f_x\)
\(\therefore f_x = 2 x e^{2y-3z}\)
\(f_y\)
By applying the rule \(\frac{dg\left(u\right)}{dy}=\frac{dg}{du}\cdot \frac{du}{dy}\)
\(g=e^u,\quad u=2y-3z\)
\(\frac{d}{du}\left(e^u\right)=e^u\)
\(\frac{d}{dy}\left(2y-3z\right)=2\)
\(f_{y} = x^2\frac{d}{du}\left(e^u\right)\frac{d}{dy}\left(2y-3z\right)\)
\(\therefore f_y = 2x^2e^{2y-3z}\)
\(f_z\)
By applying the rule \(\frac{dg\left(u\right)}{dz}=\frac{dg}{du}\cdot \frac{du}{dz}\)
\(g=e^u,\quad u=2y-3z\)
\(\frac{d}{du}\left(e^u\right)=e^u\)
\(\frac{d}{dz}\left(2y-3z\right)=-3\)
\(f_{z} = x^2\frac{d}{du}\left(e^u\right)\frac{d}{dz}\left(2y-3z\right)\)
\(\therefore f_z = -3x^2e^{2y-3z}\)
\(f_{yz}\)
\(f_y = 2x^2e^{2y-3z}\)
By applying the rule \(\frac{dg\left(u\right)}{dz}=\frac{dg}{du}\cdot \frac{du}{dz}\)
\(g=e^u,\quad u=2y-3z\)
\(\frac{d}{du}\left(e^u\right)=e^u\)
\(\frac{d}{dz}\left(2y-3z\right)=-3\)
\(f_{yz} = 2x^2\frac{d}{du}\left(e^u\right)\frac{d}{dy}\left(2y-3z\right)\)
\(f_{yz} = (-3) \cdot 2x^2e^{2y-3z}\)
\(\therefore f_{yz} = -6x^2e^{2y-3z}\)
\(f_{zy}\)
\(f_z = -3x^2e^{2y-3z}\)
By applying the rule \(\frac{dg\left(u\right)}{dy}=\frac{dg}{du}\cdot \frac{du}{dy}\)
\(g=e^u,\quad u=2y-3z\)
\(\frac{d}{du}\left(e^u\right)=e^u\)
\(\frac{d}{dy}\left(2y-3z\right)=2\)
\(f_{zy} = -3x^2\frac{d}{du}\left(e^u\right)\frac{d}{dy}\left(2y-3z\right)\)
\(\therefore f_{zy} = -6x^2e^{2y-3z}\)
Please let me know your input!
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