# Thread: mastery test; derivative question

1. ## mastery test; derivative question

im studying for this test i have to take on monday called the mastery test. its all derivatives and im studying this practice sheet my teacher posted online but i think i found an error on one. can someone check it please.

y=x^sin(x)

find y'

y'=(ln(x))(x^sin(x))(cos(x))

this is what i got ^^^^

but on the answer sheet of the practice sheet it says that the answer is:

[cosx(lnx) + sinx(1/x)] (x^sinx)

which one is correct?

thank you

2. Originally Posted by mybrohshi5
im studying for this test i have to take on monday called the mastery test. its all derivatives and im studying this practice sheet my teacher posted online but i think i found an error on one. can someone check it please.

y=x^sin(x)

find y'

y'=(ln(x))(x^sin(x))(cos(x))

this is what i got ^^^^

but on the answer sheet of the practice sheet it says that the answer is:

[cosx(lnx) + sinx(1/x)] (x^sinx)

which one is correct?

thank you
The answer sheet is correct.

$y = x^{\sin(x)}$

$\ln(y) = \ln(x^{\sin(x)})$

$\therefore \ln(y) = \sin(x) \ln (x)$

Differentiate this implicitly using the chain rule on the LHS, and using the product rule on the RHS:

The Chain Rule:

$\frac{d}{dx} f(y) = \frac{d f(y)}{dy} \times \frac{dy}{dx}$

The Product Rule:

$\frac{d}{dx} f(x) \times g(x) = f (x) \times \frac{d g(x)}{dx} + g(x) \times \frac{d f(x)}{dx}$

3. oopps. thank you. i just saw there was an x in the power and was using the theorem or whatever it is you can use like for y=5^x ..... y'=ln5(5^x)(1) but its different when there is an x raising an x to a power haha. silly me.

thanks again

4. Originally Posted by mybrohshi5
oopps. thank you. i just saw there was an x in the power and was using the theorem or whatever it is you can use like for y=5^x ..... y'=ln5(5^x)(1) but its different when there is an x raising an x to a power haha. silly me.

thanks again
It's not exactly a theorem, but when you have $y = 5^x$ the derivative is obtained using the same procedure Mush used.

It's best not to remember some formula when tackling these types of derivatives but to remember the procedure of taking the natural log of both sides, differentiating both sides (using the product rule or chain rule if necessary) and then rearranging to find $\frac{dy}{dx}$