# Thread: find y' if y=(sinx)^2 + 2^(sinx)

1. ## find y' if y=(sinx)^2 + 2^(sinx)

find y' if $y=(sinx)^2 + 2^{sinx}$

I don't really know which definitions/properties to use in this instance, but what I've tried is:

Dx $a^{x} = a^{x}\ln{a}$

First, I separated the two terms and took the derivative of both

Dx $(sinx)^2 = (sinx)^2\ln{sinx}$

Dx $2^{sinx} = 2(sinx)(cosx)$ (Chain Rule)

Giving: $(sinx)^2\ln{sinx} + 2(sinx)(cosx)$

Any chance that I did this right? If not, am I using the wrong properties for the first term? Thanks!!

2. Originally Posted by littlejodo
find y' if $y=(sinx)^2 + 2^{sinx}$

I don't really know which definitions/properties to use in this instance, but what I've tried is:

Dx $a^{x} = a^{x}\ln{a}$

First, I separated the two terms and took the derivative of both

Dx $(sinx)^2 = (sinx)^2\ln{sinx}$

Dx $2^{sinx} = 2(sinx)(cosx)$ (Chain Rule)

Giving: $(sinx)^2\ln{sinx} + 2(sinx)(cosx)$

Any chance that I did this right? If not, am I using the wrong properties for the first term? Thanks!!
Okay, for the first term, $(sinx)^2$, you would simply use the chain rule and you would get
$2(sinx)(cosx)$
The take the derivative of the second term using
$a^{x} = a^{x}\ln{a}$
and you would get as a final answer
$cosx(2(sinx)+2^{sinx}ln2)$

3. It looks like you must have had a cosx in both terms in order to be able to factor it out. Is this because in the second term after getting 2^(sinx)ln2 you have to get the derivative of the exponent, sinx, which is cosx?

Thanks!!!!

4. Originally Posted by littlejodo
It looks like you must have had a cosx in both terms in order to be able to factor it out. Is this because in the second term after getting 2^(sinx)ln2 you have to get the derivative of the exponent, sinx, which is cosx?

Thanks!!!!
yes, this is due to the chain rule

for $a > 0$

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

so, by the chain rule, $\frac d{dx}a^u = u'a^u \ln a$

where $u$ is a function of $x$

This follows from $\frac d{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$ (the Chain Rule)

in the second rule i gave you, $f(x) = a^x$ and $g(x) = u$