What is (\partial)/(\partial x)((-x^4+7x^2+3y)^6) ?

The answer to (\partial)/(\partial x)((-x^4+7x^2+3y)^6) is 6(-x^4+7x^2+3y)^5(-4x^3+14x)

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(\partial)/(\partial x)((-x^4+7x^2+3y)^6)

\frac{\partial}{\partial x} ((- x^{4} + 7 x^{2} + 3 y)^{6})

6 (- x^{4} + 7 x^{2} + 3 y)^{5} (- 4 x^{3} + 14 x)

Solution steps

\frac{\partial}{\partial x} ((- x^{4} + 7 x^{2} + 3 y)^{6})

Treat

y

as a constant

Apply the chain rule:

6 (- x^{4} + 7 x^{2} + 3 y)^{5} \frac{\partial}{\partial x} (- x^{4} + 7 x^{2} + 3 y)

= 6 (- x^{4} + 7 x^{2} + 3 y)^{5} \frac{\partial}{\partial x} (- x^{4} + 7 x^{2} + 3 y)

\frac{\partial}{\partial x} (- x^{4} + 7 x^{2} + 3 y) = - 4 x^{3} + 14 x

= 6 (- x^{4} + 7 x^{2} + 3 y)^{5} (- 4 x^{3} + 14 x)

(y - x y + x - 1) d x + x^{2} d y = 0

\int \frac{x + 49}{x} d x

\int \frac{x ^{3} ( 1 )}{1 - x ^{8}} d x

\frac{d}{d t} (- e^{- t} sin (t) + e^{- t} cos (t))

\frac{2 x ^{2}}{5}

What is (\partial)/(\partial x)((-x^4+7x^2+3y)^6) ?
The answer to (\partial)/(\partial x)((-x^4+7x^2+3y)^6) is 6(-x^4+7x^2+3y)^5(-4x^3+14x)