MHB Derivative of function with a natural log in the exponent

[tmt1]

Supposing we have $f(x) = {2}^{lnx}$, how would we find $f'(x)$?

 

[kaliprasad]

Supposing we have $f(x) = {2}^{lnx}$, how would we find $f'(x)$?

you can take log on both sides to get

ln f(x) = ln x ln 2
and differentiate both sides

alternatively

$f(x) = 2^{ln x} =e^{(ln 2) ln x} = x^{ln 2}$ and now you can differentiate

 

[tmt1]

you can take log on both sides to get

ln f(x) = ln x ln 2
and differentiate both sides

alternatively

$f(x) = 2^{ln x} =e^{(ln 2) ln x} = x^{ln 2}$ and now you can differentiate

So the answer would be ${2}^{lnx} (\frac{1}{2} lnx + \frac{1}{x} ln2)$ ?

 

[kaliprasad]

So the answer would be ${2}^{lnx} (\frac{1}{2} lnx + \frac{1}{x} ln2)$ ?

kindly show the steps

 

[tmt1]

kindly show the steps

Ok, if we take your second expression, with seems easier, we have ${x}^{ln2}$.

So, if we differentiate that, do we get something like: $ln2({x}^{ln2 - 1}) * \frac{1}{2}$. I'm not sure how to properly evaluate this. The $\frac{1}{2}$ comes from the derivative of $ln2$ but this seems incorrect.

 

[kaliprasad]

Ok, if we take your second expression, with seems easier, we have ${x}^{ln2}$.

So, if we differentiate that, do we get something like: $ln2({x}^{ln2 - 1}) * \frac{1}{2}$. I'm not sure how to properly evaluate this. The $\frac{1}{2}$ comes from the derivative of $ln2$ but this seems incorrect.

No . derivative of ln x = 1/x but as ln 2 is constant derivative is zero

so we are left with $ln2({x}^{ln2 - 1})$ which is the result using $x^n$ derivative rule

 

[HallsofIvy]

It is also possible to do this without converting to "e", using the fact that \frac{a^u}{dx}= a^u ln(a) \frac{du}{dx}

Here, since u(x)= ln(x), \frac{du}{dx}= \frac{1}{x}.
\frac{d2^{ln(x)}}{dx}= \frac{2^{ln(x)}}{x}ln(2).