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Question 15.4 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 \(g(x,y,z)= x^2+y^2\).

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

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

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

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

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

Correct.

Incorrect.

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Question 15.5 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 \({\bf c}(t)=(e^t,t,\sin t)\).

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

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

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

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

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

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Incorrect.

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15.4 Section 13.4 Progress Check Question 1 and 2

Question 15.4 Section 13.4 Progress Check Question 1

Question 15.5 Section 13.4 Progress Check Question 2