What is (\partial)/(\partial c)((a^3b^4)/(c^2d^4)) ?

The answer to (\partial)/(\partial c)((a^3b^4)/(c^2d^4)) is -(2a^3b^4)/(c^3d^4)

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(\partial)/(\partial c)((a^3b^4)/(c^2d^4))

\frac{\partial}{\partial c} (\frac{a ^{3} b ^{4}}{c ^{2} d ^{4}})

- \frac{2 a ^{3} b ^{4}}{c ^{3} d ^{4}}

Solution steps

\frac{\partial}{\partial c} (\frac{a ^{3} b ^{4}}{c ^{2} d ^{4}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{a ^{3} b ^{4}}{d ^{4}} \frac{\partial}{\partial c} (\frac{1}{c ^{2}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{a ^{3} b ^{4}}{d ^{4}} \frac{\partial}{\partial c} (c^{- 2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= \frac{a ^{3} b ^{4}}{d ^{4}} (- 2 c^{- 2 - 1})

Simplify

\frac{a ^{3} b ^{4}}{d ^{4}} (- 2 c^{- 2 - 1}) : - \frac{2 a ^{3} b ^{4}}{c ^{3} d ^{4}}

= - \frac{2 a ^{3} b ^{4}}{c ^{3} d ^{4}}

f (x) = arccos (x)

\frac{d}{d x} (3 x e^{- 3 x})

\frac{\partial}{\partial x} (2 cos (4 x))

e^{e^{t}}

\int 5 sec^{2} (5 x) tan^{5} (5 x) d x

What is (\partial)/(\partial c)((a^3b^4)/(c^2d^4)) ?
The answer to (\partial)/(\partial c)((a^3b^4)/(c^2d^4)) is -(2a^3b^4)/(c^3d^4)