Derivative of x^2sin(4x) + xcos^(-2x)

TommG
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Need to find the Derivative using the chain rule

y = x2sin4(x) + xcos-2(x)

I am not sure where to start.

answer in book is
2xsin4(x) + 4x2sin3(x)cos(x) + cos-2(x) +2xcos-3(x) xsin(x)
 
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I am not sure if I did it right but I think I got the answer.

x2sin4(x)
2xsin4 took derivative of just x2
4x2sin3x derivative of just power of sin
cosx derivative of just sin

2xsin4(x)+4x2sin3(x)cos(x)

xcos-2(x)

cos-2 derivative of x
-2xcos-3(x) derivative of power of cos
-xsin(x) derivative of just cos

cos-2+2xcos-3(x)sin(x)

final answer
2xsin4(x)+4x2sin3(x)cos(x)+cos-2(x)+2xcos-3(x)xsin(x)
 
Last edited:
Yes, but your work is very strange and hard to read, and I think it makes it more difficult than it should be.

Also, tidy up your penmanship a bit. sin² doesn't mean anything.

Just some suggestions.
 
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...
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