Need to differentiate y = x^(log base5 (x)) in terms of natural log; already tried

**dokrbb** · March 4th 2013, 03:35 PM

So, I've got such a function: y = x^(log base5 (x)), which I have to differentiate in terms of ln;

I started by solving the equation y = x^(lnx/ln5) = x^(lnx - ln5) (wonder if here is my mistake)

after that ln y = ln x^(lnx - ln5)=> 1/y*y' = (lnx - ln5)'(ln x) + (ln x)'(lnx - ln5) => (1/x - 1/5)(ln x) + (1/x)(lnx - ln5) =>

y' = x^(lnx - ln5)[(1/x - 1/5)(ln x) + ((lnx - ln5)/x)], that's what I got, but putting it in webwork it consideres it as incorrect,

tell me please were is my mistake,
Thanks,

**Educated** · March 4th 2013, 10:17 PM

Originally Posted by dokrbb

So, I've got such a function: y = x^(log base5 (x)), which I have to differentiate in terms of ln;

I started by solving the equation y = x^(lnx/ln5) = x^(lnx - ln5) (wonder if here is my mistake)
Yea this is your mistake. you would only be able to do this if it was $\displaystyle{ x^{\ln \frac{x}{5}} = x^{\ln x - \ln 5}}$

after that ln y = ln x^(lnx - ln5)=> 1/y*y' = (lnx - ln5)'(ln x) + (ln x)'(lnx - ln5) => (1/x - 1/5)(ln x) + (1/x)(lnx - ln5) =>

y' = x^(lnx - ln5)[(1/x - 1/5)(ln x) + ((lnx - ln5)/x)], that's what I got, but putting it in webwork it consideres it as incorrect,

tell me please were is my mistake,
Thanks,

Carrying on from $y = x^{\frac{\ln x}{\ln 5}}$

$\ln y = \ln (x^{\frac{\ln x}{\ln 5}})$

$\ln y = \frac{\ln x}{\ln 5} \cdot \ln (x)$

Now just do some implicit differentiation and use the product rule on the right hand side to get your answer.

**dokrbb** · March 5th 2013, 04:48 AM

so, I assume the right answer would be

y'/y = [((1/x)ln 5 - ln x (1/5))/(ln5)^(2)]*ln x + (lnx/ln5)*(1/x) =>

y' = (lnx/ln5)*ln(x) {[((1/x)ln 5 - ln x (1/5))/(ln5)^(2)]*ln x + (lnx/ln5)*(1/x)}

y' = (lnx/ln5)*ln(x){[((1/x)ln 5 - ln x (1/5))/(ln5)^(2)]*ln x + ((x lnx/ln5)}, there is a way to contract it a little bit or its better to leave it like this?

**dokrbb** · March 5th 2013, 06:34 AM

oh, let me just continue, y' = (lnx/ln5)*ln(x){[((1/x)ln 5 - ln x (1/5))/(ln5)^(2)]*ln x + ((x lnx/ln5)}, =>

y' = (lnx/ln5)*ln(x){[ (ln 5/x) - (ln x /5)/(ln5)^(2)]*ln x + ((x lnx/ln5)}, =>

y' = (lnx/ln5)*ln(x) [((lnx/((ln5)^(2))*(ln5/x)-(lnx/5)) + ((x lnx/ln5)]

does that makes sense?

**hollywood** · March 5th 2013, 06:53 AM

It's hard to read your equations, so I'll try to translate them:

$\frac{y'}{y} = [((\frac{1}{x})\ln{5} - (\ln{x}) (\frac{1}{5}))/(\ln{5})^2]*\ln{x} + (\frac{\ln{x}}{\ln{5}})*(\frac{1}{x}) \rightarrow$

$y' = (\frac{\ln{x}}{\ln{5}})*\ln{x} {[((\frac{1}{x})\ln{5} - (\ln{x}) (\frac{1}{5}))/(\ln{5})^2]*\ln{x} + (\frac{\ln{x}}{\ln{5}})*(\frac{1}{x})}$

$y' = (\frac{\ln{x}}{\ln{5}})*\ln{x}{[((\frac{1}{x})\ln{5} - (\ln{x}) (\frac{1}{5}))/(\ln{5})^2]*\ln{x} + (x \frac{\ln{x}}{\ln{5}})}$

Here's how I would have done it:

$\ln{y}=\frac{1}{\ln{5}}(\ln{x})^2$

$\frac{y'}{y}=\frac{1}{\ln{5}} (2\ln{x})\left(\frac{1}{x}\right)$

$y'=\frac{1}{\ln{5}} (2\ln{x})\left(\frac{1}{x}\right) (x^\frac{\ln{x}}{\ln{5}})$

$y'=\frac{2x^{\frac{\ln{x}}{\ln{5}}-1}\ln{x}}{\ln{5}}$

- Hollywood

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