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

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

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

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

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

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

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

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

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

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 $\displaystyle y=(sinx)^2 + 2^{sinx}$

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

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

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

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

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

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

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, $\displaystyle (sinx)^2$, you would simply use the chain rule and you would get
$\displaystyle 2(sinx)(cosx)$
The take the derivative of the second term using
$\displaystyle a^{x} = a^{x}\ln{a}$
and you would get as a final answer
$\displaystyle cosx(2(sinx)+2^{sinx}ln2)$

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 $\displaystyle a > 0$

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

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

where $\displaystyle u$ is a function of $\displaystyle x$

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

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