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

The answer to (\partial)/(\partial x)(((3))/((x+y)^3)) is -9/((x+y)^4)

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

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

- \frac{9}{( x + y ) ^{4}}

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

= 3 (- \frac{3}{( x + y ) ^{4}} \cdot 1)

Simplify

3 (- \frac{3}{( x + y ) ^{4}} \cdot 1) : - \frac{9}{( x + y ) ^{4}}

= - \frac{9}{( x + y ) ^{4}}

\int sec^{3} (x) tan (x) d x

x \to (- \frac{π}{2}) - lim (sec (x))

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

7 x + 1 (9.8)

f^{^{'}} (x) = e^{x^{2}}

What is (\partial)/(\partial x)(((3))/((x+y)^3)) ?
The answer to (\partial)/(\partial x)(((3))/((x+y)^3)) is -9/((x+y)^4)