# Problem of Calculus: derivative, i guess logarithmic differentiation

1. Jun 9, 2012

### BlaZz

1. The problem statement, all variables and given/known data

2. Relevant equations

My question is how I can solve the equation I gave above.

Should I use logarithmic differentiation?

Because I think that the logarithmic differentiation is used when y = (the equation) but my problem is f (x) = (the equation).

3. The attempt at a solution

2. Jun 9, 2012

### algebrat

try using (f+g)'=f'+g', then use logarithmic differentiation for f and g separately.

In your link to worked example, first option is wrong because you never took derivative in sense of a^(5x-1) for instance. Second option is wrong because Log does not distrubute across addition like that.

3. Jun 10, 2012

### BlaZz

What about this possible solution to the problem:($x^{(5x-1)}$$\frac{5x-1}{x}$+5lnx) - 2x$e^{(11-x^2)}$

4. Jun 10, 2012

### SammyS

Staff Emeritus
Hello BlaZz. Welcome to PF !

It really helps us, if you make your images visible in your post.

Show us how you got that.

The $\displaystyle -2x\,e^{11-x^2}$ is the derivative of $\displaystyle e^{11-x^2}.$

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Last edited: Jun 10, 2012
5. Jun 10, 2012

### BlaZz

I attach the procedure for which i got the result, i need that someone say me if i am right or wrong.

Thanks to all the members that comment in my thread and try to help me.

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6. Jun 10, 2012

### SammyS

Staff Emeritus

Yes, that answer looks fine !

7. Jun 12, 2012

### SammyS

Staff Emeritus
So, I gave the x5x-1 term to WolframAlpha to differentiate.

WolframAlpha did not use logarithmic differentiation. Instead, it used the following scheme.
Suppose we have a function that can be expressed in the following way,

$\displaystyle f(x)=g\left(u(x),\,v(x)\right)$

then the derivative of f(x) can be obtained as follows.

$\displaystyle \frac{d}{dx}f(x)=\frac{d}{dx}g\left(u(x),\,v(x) \right)=\frac{\partial g(u,v)}{\partial u}\frac{d\,u(x)}{dx}+\frac{\partial g(u,v)}{\partial v}\frac{d\,v(x)}{dx}$
For the case of $\displaystyle f(x)=x^{5x-1}\,,$ we have $\displaystyle g(u,\,v)=u^v\,,$ where $\displaystyle u(x)=x\,,$ and $\displaystyle u(x)=5x-1\,.$

∂g/∂u treats the function g, as if the base of the expression, in this case x by itself, is the part with the variable, and treats the exponent as being constant, as is the case of a power function.

∂g/∂v treats the exponent of the expression, in this case x by itself, as being the part with the variable, and treats the base as being constant, as is the case of an exponential function.