Question

f(x) = x^3*sin(x+sqrt(x+cos2x)

I get f'(x) = 3x^2*sin(x+sqrt(x+cos2x))+x^3*cos(x+(cos2x)^(-1/2))*(1+(-sin(2x)))*2....

Can someone look at this and tell me if it's right? Test tomorrow need to make sure I understand this!

Answer 1

f(x) = x^3*sin(x+sqrt(x+cos2x)) sorry

Answer 2

Let's see...
\[\small \begin{align*}
f(x)&=x^3\sin(x+\sqrt{x+\cos2x})\\\\
f'(x)&=3x^2\sin(x+\sqrt{x+\cos2x})+x^3\cos(x+\sqrt{x+\cos2x})\frac{d}{dx}[x+\sqrt{x+\cos2x}]\\\\
&=\cdots+x^3\cos(x+\sqrt{x+\cos2x})\left(1+\frac{1}{2\sqrt{x+\cos2x}}\frac{d}{dx}[x+\cos2x]\right)\\\\
&=\cdots+x^3\cos(x+\sqrt{x+\cos2x})\left(1+\frac{1}{2\sqrt{x+\cos2x}}\left(1-\sin2x\frac{d}{dx}[2x]\right)\right)\\\\
&=\cdots+x^3\cos(x+\sqrt{x+\cos2x})\left(1+\frac{1}{2\sqrt{x+\cos2x}}\left(1-2\sin2x\right)\right)
\end{align*}\]

Answer 3

Looks like my answer is the same to me, right? Except I didn't include the whole chain rule in the same (...), but I'll note that!

Answer 4

Perhaps you have it right on paper, but what you typed is
\[3x^2\sin(x+\sqrt{x+\cos2x})+2x^3\cos(x+(\cos2x)^{-1/2})(1-\sin2x)\]
which is not correct.

Answer 5

Yeah I screwed up typing it in a couple places I think, thanks for the help!