What is (\partial)/(\partial x)(7/(7x+8y)) ?

The answer to (\partial)/(\partial x)(7/(7x+8y)) is -(49)/((7x+8y)^2)

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(\partial)/(\partial x)(7/(7x+8y))

\frac{\partial}{\partial x} (\frac{7}{7 x + 8 y})

- \frac{49}{( 7 x + 8 y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{7}{7 x + 8 y})

Treat

y

as a constant

Take the constant out:

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

= 7 \frac{\partial}{\partial x} (\frac{1}{7 x + 8 y})

Apply exponent rule:

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

= 7 \frac{\partial}{\partial x} ((7 x + 8 y)^{- 1})

Apply the chain rule:

- \frac{1}{( 7 x + 8 y ) ^{2}} \frac{\partial}{\partial x} (7 x + 8 y)

= - \frac{1}{( 7 x + 8 y ) ^{2}} \frac{\partial}{\partial x} (7 x + 8 y)

\frac{\partial}{\partial x} (7 x + 8 y) = 7

= 7 (- \frac{1}{( 7 x + 8 y ) ^{2}} \cdot 7)

Simplify

7 (- \frac{1}{( 7 x + 8 y ) ^{2}} \cdot 7) : - \frac{49}{( 7 x + 8 y ) ^{2}}

= - \frac{49}{( 7 x + 8 y ) ^{2}}

\int_{- \infty}^{6} r \cdot e^{\frac{r}{3}} d r

x \to 0 + lim (x ln (2 x))

\frac{d y}{d x}, y = 5 x^{2}

f (x) = \frac{cos ( x ) - 4}{2 sin ( x )}

\frac{1}{3} t^{3} + 7 t^{2} + 50 t

What is (\partial)/(\partial x)(7/(7x+8y)) ?
The answer to (\partial)/(\partial x)(7/(7x+8y)) is -(49)/((7x+8y)^2)