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Exercise 8.8 · Q10

Q.If g(x,y)=3x2−5y+2y2, x(t)=etg(x,y)=3x^2-5y+2y^2,\ x(t)=e^t and y(t)=cos⁡ty(t)=\cos t, then dgdt\dfrac{dg}{dt} is equal to

(1) 6e2t+5sin⁡t−4cos⁡tsin⁡t6e^{2t}+5\sin t-4\cos t\sin t
(2) 6e2t−5sin⁡t+4cos⁡tsin⁡t6e^{2t}-5\sin t+4\cos t\sin t
(3) 3e2t+5sin⁡t+4cos⁡tsin⁡t3e^{2t}+5\sin t+4\cos t\sin t
(4) 3e2t−5sin⁡t+4cos⁡tsin⁡t3e^{2t}-5\sin t+4\cos t\sin t
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Apply the chain rule to g(x,y)=3x2−5y+2y2g(x,y)=3x^2-5y+2y^2 with x=et, y=cos⁡tx=e^t,\,y=\cos t, then substitute and simplify.

Step 1. Partial derivatives of gg. gx=6xg_x=6x. gy=−5+4y\quad g_y=-5+4y.

Step 2. Derivatives of x,yx,y w.r.t. tt. dxdt=et\dfrac{dx}{dt}=e^t. dydt=−sin⁡t\quad\dfrac{dy}{dt}=-\sin t.

Step 3. Combine via the chain rule.

dgdt=gxdxdt+gydydt=6x⋅et+(−5+4y)(−sin⁡t)=6xet+(5−4y)sin⁡t.\frac{dg}{dt} = g_x\frac{dx}{dt}+g_y\frac{dy}{dt} = 6x\cdot e^t + (-5+4y)(-\sin t) = 6xe^t + (5-4y)\sin t. …

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