Help with derivative of $y=x^2\sin^5x+x\cos^{-5}x$ 
Find $y^{\prime}$ of $y=x^2\sin^5x+x\cos^{-5}x$ 

My try: 
$\dfrac{d}{dx}(x^2\sin^5x)=x^2(-5\sin^4x)+(2x\sin^5x)$  
$\dfrac{d}{dx}(x\cos^{-5}x)=x(-5\cos^{-6}x)+1(\cos^{-5}x)$
This doesn't seem right. Can you please show how to do it correctly? 
The answer is $y^{\prime}=x\sin^4(x) ( 2 \sin(x)+5x\cos(x)) +(5x\tan(x)+1)\sec^5x$


$$y=x^2\sin^5x+x\cos^{-5}x$$

When taking the derivative of each summand of $y$, you left out a component of the chain rule for each of the trigonometric functions: $$[g(x)]^n = n[g(x)]^{n - 1} \cdot \color{blue}{g'(x)}$$ See the text highlighted in blue:
$$\dfrac{d}{dx}(x^2\sin^5x)=x^2(5\sin^4x){\bf\color{blue}{ \cos x}}+(2x\sin^5x)$$
$$\dfrac{d}{dx}(x\cos^{-5}x)=x(-5\cos^{-6}(x){\bf \color{blue}{(-\sin x)}})+1(\cos^{-5}x)$$
$$y' = 5x^2\sin^4x\cos x+ 2x\sin^5x +5x\cos^{-6}x\sin x + \cos^{-5}x$$
