3 questions on partial derivatives

**lllll** · March 27th 2008, 11:22 PM

Find the first partial derivatives of the function:

1) $f(x,y) = x^y$

2) $f(x,y) = \int_{y}^{x} cos(t^2) dt$

3) $u = x^{\frac{y}{z}}$

Solutions

1) $\frac{\partial f}{\partial x} = yx^{y-1}$ , $\frac{\partial f}{\partial y} = ln(x)$

2) I have no idea

3) $\frac{\partial u}{\partial x} = \frac{y}{z}x^{\frac{y}{z}-1}$ , $\frac{\partial u}{\partial y} = \frac{ln(x)}{z} = x^{\frac{1}{z}}$ , $\frac{\partial u}{\partial z} = \frac{yln(x)}{z^2} = x^{\frac{y}{z^2}}$

are these correct, and any idea on how to solve 2 would be greatly appreciated.

**TheEmptySet** · March 28th 2008, 12:15 AM

$ f(x,y) = \int_{y}^{x} cos(t^2) dt $

by FTC we get...

$\frac{\partial{f}}{\partial{x}}=\cos(x^2)$

for y we need to rewrite the integtal so that the FTC applies

$ f(x,y) = \int_{y}^{x} cos(t^2) dt=-\int_{x}^{y} \cos(t^2)dt $

Now

$\frac{\partial{f}}{\partial{y}}=-cos(y^2)$

**Peritus** · March 28th 2008, 12:16 AM

1. $\frac{{\partial f}} {{\partial y}} = x^y \ln x $

3. $\frac{{\partial u}} {{\partial y}} = \frac{1} {z}x^{\frac{y} {z}} \ln x$

$ \frac{{\partial u}} {{\partial z}} = - \frac{y} {{z^2 }}x^{\frac{y} {z}} \ln x $

**TheEmptySet** · March 28th 2008, 12:19 AM

$f(x,y)=x^y \iff ln(f(x,y))=y \ln(x)$ taking the derivative we get

$\frac{1}{f(x,y)}\frac{\partial{f}}{\partial{y}}=\l n(x)$

so

$\frac{\partial{f}}{\partial{y}}=f(x,y) \cdot \ln(x) =x^y \ln(x)$

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