# Differential of a^x^b

1. May 26, 2015

• Member warned about posting with no effort
1. The problem statement, all variables and given/known data
Find the differential of axb

2. Relevant equations

3. The attempt at a solution
Really not sure where to start, honestly. Thanks in advance :)

2. May 26, 2015

### SammyS

Staff Emeritus
Show some effort at understanding the problem.

How does a person find a differential?

3. May 26, 2015

In the picture is what I have so far.
Not sure if correct or not.

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4. May 26, 2015

### SammyS

Staff Emeritus

It looks like you let $\displaystyle\ y=a^{\displaystyle x^b}\,,\$ then found the derivative, $\displaystyle\ \frac{dy}{dx}\$.

So, now to get the differential.

5. May 26, 2015

### Ray Vickson

Perhaps he/she was being asked to find the derivative, but due to a language issue it came out as "differential".

6. May 26, 2015

### tommyxu3

As I also thought they are similar (derivative and differential), or maybe I'm wrong...?
It seems correct and I have a suggestion that you can differentiate on $x$ instead of $y$ while the process is similar (just a little easy).

7. May 26, 2015

### SammyS

Staff Emeritus
Yes. That's an excellent point!

When ninjaduck did implicit differentiation, he took the derivative with respect to y, rather than the customary method of taking the derivative with respect to x .

8. May 27, 2015

### HallsofIvy

Yes, if $y= a^{x^b}$ then $\frac{dy}{dx}= ba^{x^b}ln(a)x^{b- 1}$.

You used "logarithmic differentiation". If you do that with just $y= a^x$ then $ln(y)= x ln(a)$ so that $y'/y= ln(a)$, $y'= y ln(a)= ln(a) a^x$. I find it simpler to use that generally.

Here, we use the chain rule with $u(x)= x^b$ so that $y= a^u$. Then $y'= ln(a) a^u (u')$. Since $u= x^b$, $u'= bx^{b- 1}$. Putting those together, the derivative of $y= a^{x^b}$ is $dy/dx= ln(a)\left(a^{x^b}\right)\left(bx^{b-1}\right)$, just what you have.