0
$\begingroup$

${ x }^{ 2 }(4\sin { x+\log { x\cos { x) }}}$ ?

I tried and got the answer as $8x\cdot \cos { x }$ which is wrong.

  • 1
    It's a product, right? Start with the product rule. But note -- the inner expression is a sum two terms, one of which is a product. – 2017-02-26

1 Answers 1

4

Use the product rule, which says the derivative of $f(x)g(x)$ is $f'(x)g(x)+f(x)g'(x)$.

Note that the derivative of $x^2(4 \sin x+ \log x \cos x)$ is $$2x (4 \sin x+\log x \cos x)+x^2\left(4 \cos x+ \frac{\cos x}{x}-\log x \sin x\right) $$ I think you can simplify this. This uses that $$\frac{\mathrm{d}}{\mathrm{d}x} \cos x=-\sin x$$$$\frac{\mathrm{d}}{\mathrm{d}x} \sin x=\cos x$$$$\frac{\mathrm{d}}{\mathrm{d}x} \log x=\frac{1}{x}$$ From your answer, I'm guessing you only calculated $2x (4 \sin x)$.