What is (\partial)/(\partial x)(cos^7(x^9y^8)) ?

The answer to (\partial)/(\partial x)(cos^7(x^9y^8)) is -63x^8y^8cos^6(x^9y^8)sin(x^9y^8)

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(\partial)/(\partial x)(cos^7(x^9y^8))

\frac{\partial}{\partial x} (cos^{7} (x^{9} y^{8}))

- 63 x^{8} y^{8} cos^{6} (x^{9} y^{8}) sin (x^{9} y^{8})

Solution steps

\frac{\partial}{\partial x} (cos^{7} (x^{9} y^{8}))

Treat

y

as a constant

Apply the chain rule:

7 (cos (x^{9} y^{8}))^{6} \frac{\partial}{\partial x} (cos (x^{9} y^{8}))

= 7 (cos (x^{9} y^{8}))^{6} \frac{\partial}{\partial x} (cos (x^{9} y^{8}))

\frac{\partial}{\partial x} (cos (x^{9} y^{8})) = - sin (x^{9} y^{8}) \cdot 9 y^{8} x^{8}

= 7 (cos (x^{9} y^{8}))^{6} (- sin (x^{9} y^{8}) \cdot 9 y^{8} x^{8})

Simplify

= - 63 x^{8} y^{8} cos^{6} (x^{9} y^{8}) sin (x^{9} y^{8})

\frac{1}{e ^{x}}

(y^{3} - 1) e^{x} + 3 y^{2} (e^{x} + 1) \frac{d y}{d x} = 0, y (0) = 2

y^{^{''}} - 2 y^{^{'}} - 12 y = 0

\frac{d}{d x} (\frac{5}{3} x^{3})

(- 2, 5), (4, - 4)

What is (\partial)/(\partial x)(cos^7(x^9y^8)) ?
The answer to (\partial)/(\partial x)(cos^7(x^9y^8)) is -63x^8y^8cos^6(x^9y^8)sin(x^9y^8)