Gradient of function of functions

**maximus101** · February 10th 2011, 05:43 AM

if we have functions f and g, and we let h:=sqrt((f^2) + (e^-fg)) calculate ∇h in terms of f,g,∇f, ∇g?

**Prove It** · February 10th 2011, 05:59 AM

I can't read your function for $\displaystyle h$ .

Is it $\displaystyle h = \sqrt{f^2 + e^{-fg}}$ or $\displaystyle h = \sqrt{f^2 + e^{-f}g}$ ?

**maximus101** · February 10th 2011, 06:02 AM

Hi, sorry it's $\displaystyle h = \sqrt{f^2 + e^{-fg}}$

**HallsofIvy** · February 10th 2011, 07:07 AM

The symbol you use is $\nabla$ which means the gradient vector. Are we to assume these are vector functions? Is so, what is meant by the exponential or the product of two vectors?

If you simply meant "the derivative" and these are all real valued functions of a single variable, then use the chain rule.
$\frac{\partial h}{\partial f}= \frac{1}{2}\left(f^2+ e^{-fg}\right)^{-1/2}(2f- ge^{-fg})$ ,
$\frac{\partial h}{\partial g}= \frac{1}{2}\left(f^2+ e^{-fg}\right)^{-1/2}(-fe^{-fg})$
and then
$\frac{d h}{d x}= \frac{\partial h}{\partial f}\frac{df}{dx}+ \frac{\partial h}{\partial g}\frac{dg}{dx}$

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