Help with Derivatives of f(g(h(x))) - I'm Stuck!

  • Thread starter Thread starter hawkblader
  • Start date Start date
  • Tags Tags
    Derivatives
hawkblader
Messages
4
Reaction score
0

Homework Statement



Let f(x) = sqrt(sin(e^(x^4*sinx)))

Find f '(x)

I tried this many times and it's really frustrating!

I made it a composition function : f(g(h(x)))

but I'm not getting the answer at all.

At the end, I found :
1/2(sin(e^((x^4)(sinx))))^(-1/2) (cos(e^(x^4 sinx))) (e^(x^4 sinx)) (4x^3 cosx)

But it's wrong.

Could someone please tell me where I went wrong or how to solve this please.
 
Physics news on Phys.org
hawkblader said:
1/2(sin(e^((x^4)(sinx))))^(-1/2) (cos(e^(x^4 sinx))) (e^(x^4 sinx)) (4x^3 cosx+4*3x^2 sinx)

You left out the blue part.

ehild
 
Could you please tell me how you got the blue part?

I put it in and it's still wrong :(
 
hawkblader said:
Could you please tell me how you got the blue part?

I put it in and it's still wrong :(
Product rule.
 
hawkblader said:
1/2(sin(e^((x^4)(sinx))))^(-1/2) (cos(e^(x^4 sinx))) (e^(x^4 sinx)) (4x^3 cosx)

Sorry, I also was mistaken, the red part is wrong, as you have to apply the product rule on x4sinx.

ehild
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top