derivative?

[raven2k7]

hi,i am uploaded the function that i need to get the derivative from. which shows my working. the problem i have is that i dont know how they got the square root sen^2x in the last part. the photo is attached

 

[pka]

The answer you have is correct. However if you had noticed that \(f(x)=\sqrt[5]{\sin^3(x)}=\sin^{\frac{3}{5}}(x)\)​

Would have made the work easier.​

 

[Dr.Peterson]

If you use the method you show, then the final step can be to write [MATH]\left(\sin^3(x)\right)^{-4/5}\sin^2(x)[/MATH] as [MATH](\sin(x))^{3\cdot-4/5}(\sin(x))^2 = (\sin(x))^{-12/5 + 10/5}= (\sin(x))^{-2/5}[/MATH]. Then rewrite as a radical.

But pka's method is much easier.

 

[raven2k7]

The answer you have is correct. However if you had noticed that f(x)=5√sin3(x)=sin35(x)f(x)=sin3⁡(x)5=sin35⁡(x)​

Would have made the work easier.​

yea, that's the answer that's in the book. But what I don't understand is in the part where it says 3sen^2x*cosx , what happened to that (3sen^2x) was it cancelled out ? and how and why?

 

[Dr.Peterson]

yea, that's the answer that's in the book. But what I don't understand is in the part where it says 3sen^2x*cosx , what happened to that (3sen^2x) was it cancelled out ? and how and why?

It was not cancelled; it was combined with the other power of the sine. See my post #3, where I ignored the 1/5 and the 3, and focused on the sines.

 

[raven2k7]

oh yea! you are right! i guess thats some other rule, cause i cant remember learning that one.