Problem of Calculus: derivative, i guess logarithmic differentiation

[BlaZz]

Homework Statement

Image of the problem: http://prntscr.com/addkf

Homework 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).

The Attempt at a Solution

Possible solutions: http://prntscr.com/addwu

 

[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.

 

[BlaZz]

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

 

[SammyS]

Homework Statement

Image of the problem: http://prntscr.com/addkf

Homework 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).

The Attempt at a Solution

Possible solutions: http://prntscr.com/addwu

Hello BlaZz. Welcome to PF !

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

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

Show us how you got that.

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

 

[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.

 

[SammyS]

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.

Yes, that answer looks fine !

 

[SammyS]

So, I gave the x^5x-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.