Differentiation - One question per page

15.5 Section 13.4 Progress Check Question 1, 2 and 3

Question 15.6 Section 13.4 Progress Check Question 1

Use the product rule to find the derivative \({\bf D}h\) of \(h(x,y,z) = f(x,y,z) \cdot g(x,y,z) \) where \(f(x,y,z)= xy+e^z\) and \(g(x,y,z)= x^2+y^2\).

\(\ \ \displaystyle {\bf D}h(x,y,z)= (x^2+y^2)(y,x,e^z)+(xy+e^z)(2x,2y,0)= x^3+y^3+3x^2y+3xy^2+2xe^z+2ye^z+ x^2e^z+y^2e^z)\)

\(\ \ \displaystyle {\bf D}h(x,y,z)= (2x+2y)(y,x,e^z)+(y+x+e^z)(2x,2y,0)= (2x^2+4xy+2y^2+2xe^z,2x^2+4xy+2y^2+2ye^z,2xe^z+2ye^z)\)

\(\ \ \displaystyle {\bf D}h(x,y,z)= (x^2+y^2)(y,x,e^z)+(xy+e^z)(2x,2y,0)= (3x^2y+y^3+2xe^z,x^3+3xy^2+2ye^z,x^2e^z+y^2e^z)\)

\(\ \ \displaystyle {\bf D}h(x,y,z)= (2x+2y)(y,x,e^z)+(y+x+e^z)(2x,2y,0)= 4x^2+8xy+4y^2+4xe^z+4ye^z\)

\(\ \ \displaystyle {\bf D}h(x,y,z)= (xy+e^z)(x^2+y^2)+(y,x,e^z)(2x,2y,0)= x^3y+xy^3+4xy+x^2e^z+y^2e^z\)

Correct.

Incorrect.

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Question 15.7 Section 13.4 Progress Check Question 2

Use the chain rule to find the derivative \(\displaystyle \frac{dh}{dt}\) of \(h(t) = f({\bf c}(t)) \) where \(f(x,y,z)= xe^{yz}\) and \({\bf c}(t)=(e^t,t,\sin t)\).

\(\ \ \displaystyle \frac{dh}{dt}= (e^{t\sin t},e^t e^{t\sin t}\sin t, t e^t e^{t\sin t})\)

\(\ \ \displaystyle \frac{dh}{dt}= (e^{t\sin t}+e^t,e^t e^{t\sin t}\sin t+1, t e^t e^{t\sin t}+\cos t)\)

\(\ \ \displaystyle \frac{dh}{dt}= e^t e^{\cos t}+ t e^t e^{\cos t}\cos t +e^t e^{\cos t}\sin t \)

\(\ \ \displaystyle \frac{dh}{dt}= e^t e^{t \sin t}+ e^t e^{t \sin t}\sin t+t e^t e^{t \sin t}\cos t\)

\(\ \ \displaystyle \frac{dh}{dt}= e^t e^{\cos t}\)

Correct.

Incorrect.

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Question 15.8 Section 13.4 Progress Check Question 3

Let \(f(x,y) = (xy,x, y^3)\) and \(g(u,v)=(u^2+1,u-v^2)\). Calculate \({\bf D} (f\circ g)(1,1)\) using the chain rule.

\(\ \ \displaystyle {\bf D} (f\circ g)(1,1) = \left[ \begin{array}{c@{\quad}c} 2&0\\ 4&0\\ 0&0 \end{array} \right] \)

\(\ \ \displaystyle {\bf D} (f\circ g)(1,1) = \left[ \begin{array}{c@{\quad}c} 2&-4\\ 2&0\\ 0&0\end{array} \right] \)

\(\ \ \displaystyle {\bf D} (f\circ g)(1,1) = \left[ \begin{array}{ccc} 2&2&0\\ 2&0&4 \end{array} \right] \)

\(\ \ \displaystyle {\bf D} (f\circ g)(1,1) = \left[ \begin{array}{c@{\quad}c} 3&-2\\ 2&0\\ 1&-2 \end{array} \right] \)

\(\ \ \displaystyle {\bf D} (f\circ g)(1,1) = \left[ \begin{array}{ccc} 3&2&1\\ -2&0&-2 \end{array} \right] \)

Correct.

Try again. Remember to evaluate \({\bf D}f\) at \(g(1,1)\).

Incorrect.