What is (\partial)/(\partial x)(3*(y+1)*(x-3)^3) ?

The answer to (\partial)/(\partial x)(3*(y+1)*(x-3)^3) is 9(y+1)(x-3)^2

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(\partial)/(\partial x)(3(y+1)(x-3)^3)

\frac{\partial}{\partial x} (3 \cdot (y + 1) \cdot (x - 3)^{3})

9 (y + 1) (x - 3)^{2}

Solution steps

\frac{\partial}{\partial x} (3 (y + 1) (x - 3)^{3})

Treat

y

as a constant

Take the constant out:

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

= 3 (y + 1) \frac{\partial}{\partial x} ((x - 3)^{3})

Apply the chain rule:

3 (x - 3)^{2} \frac{\partial}{\partial x} (x - 3)

= 3 (x - 3)^{2} \frac{\partial}{\partial x} (x - 3)

\frac{\partial}{\partial x} (x - 3) = 1

= 3 (y + 1) \cdot 3 (x - 3)^{2} \cdot 1

Simplify

= 9 (y + 1) (x - 3)^{2}

\int (x^{2} - 4)^{3} x d x

n = 1 \sum \infty \frac{- 4 ^{n}}{4 ^{n}}

\int_{0}^{\infty} e^{- 0.8 x} d x

y^{^{''}} + 8 y^{^{'}} + 32 y = 32, y (0) = 2, y^{^{'}} (0) = - 20

\int cos (x) cos (2 x) cos (3 x) d x

What is (\partial)/(\partial x)(3*(y+1)*(x-3)^3) ?
The answer to (\partial)/(\partial x)(3*(y+1)*(x-3)^3) is 9(y+1)(x-3)^2