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Calculus

Partial derivatives

Compute partial derivatives of f(x, y)f(x,\ y) and apply them to tangent planes, the chain rule and extremum problems.

Basic, Standard, Advanced: University Year 1 · 2nd semester

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Partial derivatives

fxf_x is the derivative with respect to xx treating yy as a constant; fyf_y treats xx as a constant.

f(x, y)=x2y+exy⇒fx=2xy+yexy,fy=x2+xexyf(x,\ y) = x^2y + e^{xy} \quad\Rightarrow\quad f_x = 2xy + ye^{xy},\quad f_y = x^2 + xe^{xy}

Second partial derivatives fxxf_{xx}, fxyf_{xy}, fyyf_{yy} are found the same way. If fxyf_{xy} and fyxf_{yx} are continuous, then fxy=fyxf_{xy} = f_{yx} (Schwarz's theorem).

Total differential and tangent planes

Formulas
dz=fx dx+fy dydz = f_x\,dx + f_y\,dy

The tangent plane to z=f(x, y)z = f(x,\ y) at (a, b, f(a, b))(a,\ b,\ f(a,\ b)) is

z−f(a, b)=fx(a, b)(x−a)+fy(a, b)(y−b)z - f(a,\ b) = f_x(a,\ b)(x - a) + f_y(a,\ b)(y - b)

The approximation f(a+h, b+k)≈f(a, b)+fx(a, b)h+fy(a, b)kf(a + h,\ b + k) \approx f(a,\ b) + f_x(a,\ b)h + f_y(a,\ b)k lets you estimate values such as 3.012+3.992\sqrt{3.01^2 + 3.99^2} by hand.

The chain rule

Chain rule
z=f(x, y), x=x(t), y=y(t):dzdt=fxdxdt+fydydtz = f(x,\ y),\ x = x(t),\ y = y(t):\quad \frac{dz}{dt} = f_x\frac{dx}{dt} + f_y\frac{dy}{dt}
x=x(s, t), y=y(s, t):zs=fxxs+fyys,zt=fxxt+fyytx = x(s,\ t),\ y = y(s,\ t):\quad z_s = f_xx_s + f_yy_s,\quad z_t = f_xx_t + f_yy_t

Extrema and constrained extrema

At a critical point (fx=fy=0f_x = f_y = 0), compute D=fxxfyy−fxy2D = f_{xx}f_{yy} - f_{xy}^2:

  • D>0D > 0 and fxx>0f_{xx} > 0: local minimum
  • D>0D > 0 and fxx<0f_{xx} < 0: local maximum
  • D<0D < 0: not an extremum (saddle point)

For extrema subject to g(x, y)=0g(x,\ y) = 0, use Lagrange multipliers: with F=f−λgF = f - \lambda g, solve Fx=Fy=0F_x = F_y = 0 and g=0g = 0 to find the candidates.