Derivative of Cos^3(x)*Sin(x) using Chain Rule

  • Thread starter Thread starter CloDawg
  • Start date Start date
  • Tags Tags
    Derivative
CloDawg
Messages
5
Reaction score
0
Help!
I last did this kind of work years ago in university. Can som1 please help me with the derivative.
 
Physics news on Phys.org
I think the question is ((cosx)^3)*sin(x).Am I right?The formula for d(uv)/dx=u(dv/dx)+v(du/dx).And of course I think you need only formula.Do you need answer?
 
omkar13 said:
I think the question is ((cosx)^3)*sin(x).Am I right?The formula for d(uv)/dx=u(dv/dx)+v(du/dx).And of course I think you need only formula.Do you need answer?
It is against Physics Forums policy to provide answers to posters' questions.
 
Thats it chain rule or something:


d(uv)/dx=u(dv/dx)+v(du/dx)

u=sin(x)
v=cos^3(x)

(dv/dx) = -3sin((x))cos^2(x)
(du/dx) = cos(x)

d(uv)/dx= -3sin^2(x)cos^2(x) + cos^4(x)

Is this correct?
 
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