# Thread: Differentiating Constant To Power x

1. ## Differentiating Constant To Power x

How would you differentiate something like:

$y = 2^x$ With respect to $x$?

2. $\frac{d}{dx}a^x = a^x \ln a$

Using substitution of $y=2^x$, solve $\int^1_0 \frac{2^x}{(2^x + 1)^2} \ \mathrm{d}x$

4. Originally Posted by Air

Using substitution of $y=2^x$, solve $\int^1_0 \frac{2^x}{(2^x + 1)^2} \ \mathrm{d}x$
A better substitution is $u = 2^x + 1$.

Then, $du = 2^x \ln 2~dx$.

$2^x ~dx=\frac{du}{\ln 2}$

The bounds are now from 2 to 3.

The integral becomes $\frac{1}{\ln 2} \int_2^3 \frac{du}{u^2}$

It's easy from here..

5. Originally Posted by wingless
A better substitution is $u = 2^x + 1$.

Then, $du = 2^x \ln 2~dx$.

$2^x ~dx=\frac{du}{\ln 2}$

The bounds are now from 2 to 3.

The integral becomes $\frac{1}{\ln 2} \int_2^3 \frac{du}{u^2}$

It's easy from here..
Yes, I would have done that too but they provided the substitution value. How would I go about with that?

6. Originally Posted by Air
Yes, I would have done that too but they provided the substitution value. How would I go about with that?
Only a very minor change to what Wingless did is required:

$\frac{1}{\ln 2} \int_1^2 \frac{du}{(u + 1)^2} = \frac{1}{\ln 2} \int_1^2 (u + 1)^{-2} \, du$.

7. Originally Posted by mr fantastic
Only a very minor change to what Wingless did is required:

$\frac{1}{\ln 2} \int_1^2 \frac{du}{(u + 1)^2} = \frac{1}{\ln 2} \int_1^2 (u + 1)^{-2} \, du$.
What rule would I use to integrate? Just reverse chain rule?

8. Originally Posted by Air
What rule would I use to integrate? Just reverse chain rule?
Standard form ( $n \neq -1$): $\int (au + b)^n \, du = \frac{1}{a} \, \frac{1}{(n+1)} \, (au + b)^{n+1} (+ C)$.

Or you could make another substitution: w = u + 1 .....

9. Also, how do you differentiate:

$y=2^{x-1}$.

Is it the same rule?

10. Hello,

Originally Posted by Air
Also, how do you differentiate:

$y=2^{x-1}$.

Is it the same rule?
Yes it is.

Actually, you can see it this way :

$a^{f(x)}=e^{f(x) \ln a}$

--> $\left(a^{f(x)}\right)'=\ln a \cdot f'(x) e^{f(x) \ln a}=f'(x) \ln a \cdot a^{f(x)}$

Here, $f(x)=x-1$

You could also find this result by applying the chain rule and using wingless's formula

In another way, you can see that $2^{x-1}=\frac 12 \cdot 2^x$