how to differentiate
y=x^2cosx-2xsinx-2cosx
y=cos^4x-sin^4x
y=(Inx^2)^2

Question

how to differentiate
y=x^2cosx-2xsinx-2cosx
y=cos^4x-sin^4x
y=(Inx^2)^2

anonymous · Answer

$$y=x^2\cos x-2x\sin x-2\cos x$$
Product rule for the first two terms:
$$y'=\left(2x\cos x+x^2(-\sin x)\right)-\left(2\sin x+2x\cos x\right)-2(-\sin x)\
y'=2x\cos x-x^2\sin x-2\sin x-2x\cos x+2\sin x\
y'=-x^2\sin x$$

anonymous · Answer

$$y=\cos^4x-\sin^4x$$
Apply the chain/power rules. It might help to write $\cos^4x=(\cos x)^4$ as a visual aid:
$$y'=4(\cos x)^3(-\sin x)-4(\sin x)^3(\cos x)\
y'=-4\cos^3x\sin x-4\sin^3x\cos x$$

anonymous · Answer

$$y=(\ln x^2)^2$$
Power rule and chain rule again:
$$y'=2(\ln x^2)\left(\frac{2x}{x^2}\right)=\frac{4\ln x^2}{x}=\frac{\ln x^8}{x}$$

anonymous · Answer

tnxxx how about  $$\csc^{-1}(2e^{3x})$$$${x}\cot^{-1}x+ In \sqrt{x^{2}+1}$$$$\coth^{-1}(coshx)$$$$\cot^{-1}\frac{ 2 }{ x } + \tan^{-1}\frac{ x }{ 2 } $$?