Differential of axb: Solving for a^x^b | Homework Help

[ninjaduck]

Homework Statement

Find the differential of a^xb

Homework Equations

The Attempt at a Solution

Really not sure where to start, honestly. Thanks in advance :)

 

[SammyS]

Homework Statement

Find the differential of a^xb

Homework Equations

The Attempt at a Solution

Really not sure where to start, honestly. Thanks in advance :)

Show some effort at understanding the problem.

How does a person find a differential?

 

[ninjaduck]

Show some effort at understanding the problem.

How does a person find a differential?

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

 

[SammyS]

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

That's pretty difficult to read.

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.

 

[Ray Vickson]

That's pretty difficult to read.

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.

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

 

[tommyxu3]

As I also thought they are similar (derivative and differential), or maybe I'm wrong...?

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

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).

 

[SammyS]

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).

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 .

 

[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.