# How do you differentiate y = x^2(sinx)^4 + x(cosx)^-2?

Oct 16, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 x {\left(\sin x\right)}^{4} + 4 {x}^{2} {\left(\sin x\right)}^{3} \cdot \cos x + {\left(\cos x\right)}^{-} 2 - 3 x {\left(\cos x\right)}^{-} 2 \cdot \sin x$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{2} {\left(\sin x\right)}^{4} + x {\left(\cos x\right)}^{-} 2\right)$ (differentiate both parts)
$\iff \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{2} {\left(\sin x\right)}^{4}\right) + \frac{d}{\mathrm{dx}} \left(x {\left(\cos x\right)}^{-} 2\right)$ (derivative of sum is sum of derivatives)
$\iff \frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{d}{\mathrm{dx}} \left({x}^{2}\right) \cdot {\left(\sin x\right)}^{4} + {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left({\left(\sin x\right)}^{4}\right)\right) + \frac{d}{\mathrm{dx}} \left(x {\left(\cos x\right)}^{-} 2\right)$(product rule)
$\iff \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x {\left(\sin x\right)}^{4} + {x}^{2} \cdot 4 \cdot {\left(\sin x\right)}^{3} \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right)\right) + \frac{d}{\mathrm{dx}} \left(x {\left(\cos x\right)}^{-} 2\right)$ (power rule and chain rule)
$\iff \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x {\left(\sin x\right)}^{4} + 4 {x}^{2} {\left(\sin x\right)}^{3} \cdot \cos \left(x\right)\right) + \frac{d}{\mathrm{dx}} \left(x {\left(\cos x\right)}^{-} 2\right)$ (trigonometric derivative)
$\iff \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x {\left(\sin x\right)}^{4} + 4 {x}^{2} {\left(\sin x\right)}^{3} \cdot \cos \left(x\right)\right) + \left(\frac{d}{\mathrm{dx}} \left(x\right) \cdot {\left(\cos x\right)}^{-} 2 + x \cdot \frac{d}{\mathrm{dx}} \left({\left(\cos x\right)}^{-} 3\right)\right)$ (product rule)
$\iff \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x {\left(\sin x\right)}^{4} + 4 {x}^{2} {\left(\sin x\right)}^{3} \cdot \cos \left(x\right)\right) + \left({\left(\cos x\right)}^{-} 2 + - 3 \cdot x \cdot {\left(\cos x\right)}^{-} 2 \cdot \frac{d}{\mathrm{dx}} \left(\cos x\right)\right)$ (power rule and chain rule)
$\iff \frac{\mathrm{dy}}{\mathrm{dx}} = \left(2 x {\left(\sin x\right)}^{4} + 4 {x}^{2} {\left(\sin x\right)}^{3} \cdot \cos \left(x\right)\right) + \left({\left(\cos x\right)}^{-} 2 + - 3 \cdot x \cdot {\left(\cos x\right)}^{-} 2 \cdot \sin \left(x\right)\right)$ (trigonometric derivative)

So the derivative is:
$2 x {\left(\sin x\right)}^{4} + 4 {x}^{2} {\left(\sin x\right)}^{3} \cdot \cos x + {\left(\cos x\right)}^{-} 2 - 3 x {\left(\cos x\right)}^{-} 2 \cdot \sin x$

You can try simplifying this, but I don't think there's much to simplify.