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 $y^{\prime}=x\sin^4(x) ( 2 \sin(x)+5x\cos(x)) +(5x\tan(x)+1)\sec^5x$

$\dfrac{d}{dx}(x^2\sin^5x)=x^25\sin^4x \cos x+(2x\sin^5x)$  
$\dfrac{d}{dx}(x\cos^{-5}x)=5x\cos^{-6}x \sin {x}+1(\cos^{-5}x)$
