Differentiate the function.

Question

I can't figure out how to differentiate this function. Any help would be appreciated.

y=ln(x⁹(sin(x))²)

Find y'(x)=

Answer 1

y=ln(x⁹(sin(x))²)

 

Let's use the log identity: ln(a*b) = ln(a) + ln(b)

 

y = ln(x⁹) + ln(sin(x)²)

 

Now let's use another log identity: ln(a^b) = b*ln(a)

 

y = 9ln(x) + 2ln(sin(x))

 

Can you complete it now?  For the 2ln(sin(x)), let u = sin(x) and use the chain rule.

Answer 2

Hi Jordan,

 

The answer of Surendra is a very good one. I'll offer a second approach -- let's simplify the ln expression ...

1. using the identity ln(ab) = ln(a) + ln(b), we get y = ln(x⁹) + ln(sin(x)²)
2. using the identity ln(c^d) = d*ln(c), we get y = 9*ln(x) + 2*ln(sin(x))

Now apply the derivative of ln, d(ln(x)) = (1/x)dx.

 

Can you finish it for me ?

 

Answer 3

 

y'(x)=. [1/(x^9)(sin(x)^2)] *  [9x^8 (sin(x)^2)+ 2sin(x)cos(x)* (x^9)]

 

we are using the chain rule of differentiation here.