Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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