Skip to main content
Sign in Create a free account
Varsity Tutors
Sign in

Computing Partial Derivatives - Calculus 3

  1. Given f(x,y)=x3+3x2y−y3f(x,y) = x^3 + 3x^2y - y^3, find the value of ∂3f∂x2∂y\frac{\partial^3 f}{\partial x^2 \partial y} at any point (x,y)(x,y).
    Answer: 66 We compute the third-order mixed partial derivative step by step. First: fx(x,y)=3x2+6xyf_x(x,y) = 3x^2 + 6xy. Then: fxx(x,y)=6x+6yf_{xx}(x,y) = 6x + 6y. Finally: ∂3f∂x2∂y=∂∂y[fxx]=∂∂y[6x+6y]=6\frac{\partial^3 f}{\partial x^2 \partial y} = \frac{\partial}{\partial y}[f_{xx}] = \frac{\partial}{\partial y}[6x + 6y] = 6. This is a constant, independent of the point (x,y)(x,y). Choice B incorrectly includes an x term from a calculation error, choice C represents ∂3f∂x3\frac{\partial^3 f}{\partial x^3}, choice D would occur if all cubic terms cancelled.
  2. Consider h(s,t)=sth(s,t) = s^t where s>0s > 0. Find hst(e,1)h_{st}(e,1).
    Answer: 22 For h(s,t)=sth(s,t) = s^t, we use st=etln⁡ss^t = e^{t\ln s}. Then hs(s,t)=∂∂s[etln⁡s]=etln⁡s⋅ts=st⋅ts=tst−1h_s(s,t) = \frac{\partial}{\partial s}[e^{t\ln s}] = e^{t\ln s} \cdot \frac{t}{s} = s^t \cdot \frac{t}{s} = ts^{t-1}. Now hst(s,t)=∂∂t[tst−1]=st−1+t⋅st−1ln⁡s=st−1(1+tln⁡s)h_{st}(s,t) = \frac{\partial}{\partial t}[ts^{t-1}] = s^{t-1} + t \cdot s^{t-1}\ln s = s^{t-1}(1 + t\ln s). At (e,1)(e,1): hst(e,1)=e1−1(1+1⋅ln⁡e)=e0(1+1⋅1)=1(1+1)=2h_{st}(e,1) = e^{1-1}(1 + 1 \cdot \ln e) = e^0(1 + 1 \cdot 1) = 1(1 + 1) = 2. Choice B incorrectly computes e0e^{0} as ee, choice C forgets the tln⁡st\ln s term, choice D adds instead of multiplying terms.
  3. If F(x,y,z)=xyz+x2z−y2zF(x,y,z) = xyz + x^2z - y^2z and G(x,y,z)=x2+y2+z2−6G(x,y,z) = x^2 + y^2 + z^2 - 6, find dFdx\frac{dF}{dx} along the surface G(x,y,z)=0G(x,y,z) = 0 at the point (1,2,1)(1,2,1).
    Answer: 33 First verify that (1,2,1)(1,2,1) lies on the surface: G(1,2,1)=1+4+1−6=0G(1,2,1) = 1 + 4 + 1 - 6 = 0 ✓. Since we want the derivative along the surface, we need the constrained derivative. We have Fx=yz+2xzF_x = yz + 2xz, Fz=xy+x2−y2F_z = xy + x^2 - y^2, Gx=2xG_x = 2x, and Gz=2zG_z = 2z. Using the constraint ∂z∂x=−GxGz=−2x2z=−xz\frac{\partial z}{\partial x} = -\frac{G_x}{G_z} = -\frac{2x}{2z} = -\frac{x}{z}, the constrained derivative is dFdx=Fx+Fz∂z∂x\frac{dF}{dx} = F_x + F_z\frac{\partial z}{\partial x}. At (1,2,1)(1,2,1): Fx=2(1)+2(1)(1)=4F_x = 2(1) + 2(1)(1) = 4, Fz=(1)(2)+1−4=−1F_z = (1)(2) + 1 - 4 = -1, and ∂z∂x=−11=−1\frac{\partial z}{\partial x} = -\frac{1}{1} = -1. Therefore dFdx=4+(−1)(−1)=3\frac{dF}{dx} = 4 + (-1)(-1) = 3.

Start this quiz