How do you differentiate $y = x^{2} {(sin x)}^{4} + x {(cos x)}^{- 2}$ $y = x^2(sinx)^4 + x(cosx)^-2$ ?

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How do you differentiate $y = x^{2} {(sin x)}^{4} + x {(cos x)}^{- 2}$ $y = x^2(sinx)^4 + x(cosx)^-2$ ?

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Answer 1 · 2015-10-16T18:22:17

$\frac{d y}{d x} = 2 x {(sin x)}^{4} + 4 x^{2} {(sin x)}^{3} \cdot cos x + {(cos x)}^{- 2} - 3 x {(cos x)}^{- 2} \cdot sin x$ $dy/dx = 2x(sin x)^4 + 4x^2(sin x)^3*cos x + (cos x)^-2-3x(cos x)^-2*sin x$

Explanation:

$\frac{d y}{d x} = \frac{d}{d x} (x^{2} {(sin x)}^{4} + x {(cos x)}^{- 2})$ $dy/dx = d/dx(x^2(sin x)^4 + x(cos x)^-2)$ (differentiate both parts)
$\Leftrightarrow \frac{d y}{d x} = \frac{d}{d x} (x^{2} {(sin x)}^{4}) + \frac{d}{d x} (x {(cos x)}^{- 2})$ $<=>dy/dx = d/dx(x^2(sin x)^4) + d/dx(x(cos x)^-2)$ (derivative of sum is sum of derivatives)
$\Leftrightarrow \frac{d y}{d x} = (\frac{d}{d x} (x^{2}) \cdot {(sin x)}^{4} + x^{2} \cdot \frac{d}{d x} ({(sin x)}^{4})) + \frac{d}{d x} (x {(cos x)}^{- 2})$ $<=>dy/dx = (d/dx(x^2)*(sin x)^4 + x^2*d/dx((sin x)^4)) + d/dx(x(cos x)^-2)$ (product rule)
$\Leftrightarrow \frac{d y}{d x} = (2 x {(sin x)}^{4} + x^{2} \cdot 4 \cdot {(sin x)}^{3} \cdot \frac{d}{d x} (sin x)) + \frac{d}{d x} (x {(cos x)}^{- 2})$ $<=>dy/dx = (2x(sin x)^4 + x^2*4*(sin x)^3*d/dx(sin x)) + d/dx(x(cos x)^-2)$ (power rule and chain rule)
$\Leftrightarrow \frac{d y}{d x} = (2 x {(sin x)}^{4} + 4 x^{2} {(sin x)}^{3} \cdot cos (x)) + \frac{d}{d x} (x {(cos x)}^{- 2})$ $<=>dy/dx = (2x(sin x)^4 + 4x^2(sin x)^3*cos(x)) + d/dx(x(cos x)^-2)$ (trigonometric derivative)
$\Leftrightarrow \frac{d y}{d x} = (2 x {(sin x)}^{4} + 4 x^{2} {(sin x)}^{3} \cdot cos (x)) + (\frac{d}{d x} (x) \cdot {(cos x)}^{- 2} + x \cdot \frac{d}{d x} ({(cos x)}^{- 3}))$ $<=>dy/dx = (2x(sin x)^4 + 4x^2(sin x)^3*cos(x)) + (d/dx(x)*(cos x)^-2 + x*d/dx((cos x)^-3))$ (product rule)
$\Leftrightarrow \frac{d y}{d x} = (2 x {(sin x)}^{4} + 4 x^{2} {(sin x)}^{3} \cdot cos (x)) + ({(cos x)}^{- 2} + - 3 \cdot x \cdot {(cos x)}^{- 2} \cdot \frac{d}{d x} (cos x))$ $<=>dy/dx = (2x(sin x)^4 + 4x^2(sin x)^3*cos(x)) + ((cos x)^-2 + -3*x*(cos x)^-2*d/dx(cos x))$ (power rule and chain rule)
$\Leftrightarrow \frac{d y}{d x} = (2 x {(sin x)}^{4} + 4 x^{2} {(sin x)}^{3} \cdot cos (x)) + ({(cos x)}^{- 2} + - 3 \cdot x \cdot {(cos x)}^{- 2} \cdot sin (x))$ $<=>dy/dx = (2x(sin x)^4 + 4x^2(sin x)^3*cos(x)) + ((cos x)^-2 + -3*x*(cos x)^-2*sin(x))$ (trigonometric derivative)

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

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

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How do you differentiate $y = x^{2} {(sin x)}^{4} + x {(cos x)}^{- 2}$ $y = x^2(sinx)^4 + x(cosx)^-2$ ?

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Explanation:

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