What is (\partial)/(\partial x)(3/(3x+5y)) ?

The answer to (\partial)/(\partial x)(3/(3x+5y)) is -9/((3x+5y)^2)

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(\partial)/(\partial x)(3/(3x+5y))

\frac{\partial}{\partial x} (\frac{3}{3 x + 5 y})

- \frac{9}{( 3 x + 5 y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{3}{3 x + 5 y})

Treat

y

as a constant

Take the constant out:

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

= 3 \frac{\partial}{\partial x} (\frac{1}{3 x + 5 y})

Apply exponent rule:

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

= 3 \frac{\partial}{\partial x} ((3 x + 5 y)^{- 1})

Apply the chain rule:

- \frac{1}{( 3 x + 5 y ) ^{2}} \frac{\partial}{\partial x} (3 x + 5 y)

= - \frac{1}{( 3 x + 5 y ) ^{2}} \frac{\partial}{\partial x} (3 x + 5 y)

\frac{\partial}{\partial x} (3 x + 5 y) = 3

= 3 (- \frac{1}{( 3 x + 5 y ) ^{2}} \cdot 3)

Simplify

3 (- \frac{1}{( 3 x + 5 y ) ^{2}} \cdot 3) : - \frac{9}{( 3 x + 5 y ) ^{2}}

= - \frac{9}{( 3 x + 5 y ) ^{2}}

6 x^{2} ln (x), 24 ln (x)

\frac{d y}{d x} = \frac{1}{2} x + 2

x^{2} y^{^{''}} + 3 x y^{^{'}} + 2 y = 0, y (1) = 0.2, y^{^{'}} (1) = 0.3

\frac{d}{d x} (cos (sin (tan (π) x)))

3 \cdot sin (3 t + 2)

What is (\partial)/(\partial x)(3/(3x+5y)) ?
The answer to (\partial)/(\partial x)(3/(3x+5y)) is -9/((3x+5y)^2)