What is (\partial)/(\partial x)(-15(3x+y)^4) ?

The answer to (\partial)/(\partial x)(-15(3x+y)^4) is -180(3x+y)^3

(\partial)/(\partial x)(-15(3x+y)^4)

\frac{\partial}{\partial x} (- 15 (3 x + y)^{4})

- 180 (3 x + y)^{3}

Solution steps

\frac{\partial}{\partial x} (- 15 (3 x + y)^{4})

Treat

y

as a constant

Take the constant out:

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

= - 15 \frac{\partial}{\partial x} ((3 x + y)^{4})

Apply the chain rule:

4 (3 x + y)^{3} \frac{\partial}{\partial x} (3 x + y)

= 4 (3 x + y)^{3} \frac{\partial}{\partial x} (3 x + y)

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

= - 15 \cdot 4 (3 x + y)^{3} \cdot 3

Simplify

= - 180 (3 x + y)^{3}

y^{^{'}} = (y - 1)^{2}

\frac{d}{d x} (- \frac{4 x}{( x ^{2} - 4 ) ^{2}})

\frac{d}{d x} ((\frac{x ^{2} + 6}{x ^{2} - 6})^{3})

d er i v a t i v e \frac{A}{y ^{7}} + B e^{y}

n = 1 \sum \infty \frac{3 - 2 n}{4 n}

What is (\partial)/(\partial x)(-15(3x+y)^4) ?
The answer to (\partial)/(\partial x)(-15(3x+y)^4) is -180(3x+y)^3