Thread: Derivatives of exponential functions

Originally Posted by ozgunatalay

This will require the chain rule and using logarithmic differentiation

$\displaystyle y = \frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4}{9}}}{ (x^3+6x)^{\frac{4}{11}}}$

$\displaystyle ln(y) = ln \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right)$

Since $\displaystyle ln \left(\frac{ab}{c}\right) = ln(a)+ln(b)-ln(c)$ we get


$\displaystyle ln(y) = ln \left[(x^2+2)^{\frac{3}{2}}\right] + ln \left[(x^2+9)^{\frac{4}{9}}\right] - ln \left[{(x^3+6x)^{\frac{4}{11}}}\right]$

Use the chain rule for each of them:

edit: oops no it isn't *edits*
edit2: that's better

Spoiler:

$\displaystyle \frac{1}{y} \, y' = \frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}$

$\displaystyle y' = y\, \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)$

$\displaystyle y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)$

Thank you for your solution.

I suppose you could always use the quotient rule, the product rule and the chain rule but IMO that would lead to an even bigger mess

I am learning about this new. Sometimes when I'm having difficulty solving questions. Thanks for your suggestions.

You might also want to note that there are NO exponential functions here. "Exponential functions" are functions that have the variable, x, in the exponent.

I looked again and I ask questions, I reached a different result. If I've made a mistake, correct my mistakes.

Is there anyone who can check if I did it right?

I can see no difference between your answer and e^(i*pi)'s, so I would conclude that this is correct.

EDIT: Ahh I see it now. e^(i*pi) did make a slight mistake in the last fraction:

$\displaystyle
y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x}{11(x^3+6x)^{\frac{4}{11}}}\right)
$

is supposed to be:


$\displaystyle
y' = \left(\frac{(x^2+2)^{\frac{3}{2}}(x^2+9)^{\frac{4} {9}}}{(x^3+6x)^{\frac{4}{11}}}\right) \cdot \left(\frac{3x}{(x^2+2)^{\frac{3}{2}}} + \frac{8x}{9(x^2+9)^{\frac{4}{9}}} - \frac{12x^2+24}{11(x^3+6x)^{\frac{4}{11}}}\right)
$

Where am I doing wrong?

Nowhere. Your answer is the same as the corrected one.

Sorry , I didn't check out e^(i*pi)'s answer good enough.

But your answer is correct.

Thank you for your interest in question