Find all the second partial derivatives. v=xyx−y

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lovagwb · Accepted Answer

v=xyx−y  Partially differentiate with respect to x (use the quotient rule)  vx=[dxydx]⋅(x−y)−(xy)⋅[dx−ydx](x−y)2  =y(x−y)−(xy)⋅(1−0)(x−y)2  =xy−y2−xy(x−y)2=−y2(x−y)2  Partially differentiate with respect to y (use the quotient rule)  vy=[dxydy]⋅(x−y)−(xy)⋅[dx−ydy](x−y)2  =(x)(x−y)−(xy)⋅(0−1)(x−y)2  =x2−xy+xy(x−y)2=x2(x−y)2  Partially differentiate vx with respect to x  v×=[d−y2dx]⋅(x−y)2−(−y2)⋅[d(x−y)2dx](x−y)4  =0(x−y)−(−y2)⋅2(x−y)⋅(1−0)(x−y)4  Cancel (x−y) from both the numerator and the denominator  =2y2(x−y)3  Note that: If w=yxy−x, then we can write wyy=2x2(y−x)3  But w=−v, therefore

ol3i4c5s4hr · Accepted Answer

Initial equation for 8e  T=e−2rcos⁡θ  Find first partial derivative wrt. r  Tr=−2e−2rcos⁡θ  Find second partial derivative wrt. r  Trr=4e−2rcos⁡θ  Find partial derivative wrt. theta  Tθ=−e−2rsin⁡θ  Find second partial derivative wrt theta  Tθθ=−e−2rcos⁡θ  Find the second partial derivative wrt the alternative variable  Trθ=Tθr=2e−2rsin⁡θ  Result: Trr=4e−2rcos⁡θ; Tθθ=e−2rcos⁡θ; Trθ=Tθr=2e−2rsin⁡θ

karton · Accepted Answer

v=xyx−y Find each first partial derivative, using the quotient rule. vx=y(x−y)−xy(x−y)2=−y2(x−y)2vy=x(x−y)+xy(x−y)2=x2(x−y)2 Find each second partial derivative, using the quotient rule. vxx=2y2(x−y)3vxy=−2y(x−y)2−(−2)(x−y)(−y)2(x−y)4=−2y(x−y)−2y2(x−y)3=−2xy(x−y)3vyx=2x(x−y)2−2(x−y)x2(x−y)4=2x(x−y)−2x2(x−y)3=−2xy(x−y)3vyy=2x2(x−y)3

Find all the second partial derivatives. v=\frac{xy}{x-y}

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