What is (\partial)/(\partial x)((2s-3x)/(3s+5x)) ?

The answer to (\partial)/(\partial x)((2s-3x)/(3s+5x)) is -(19s)/((3s+5x)^2)

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(\partial)/(\partial x)((2s-3x)/(3s+5x))

\frac{\partial}{\partial x} (\frac{2 s - 3 x}{3 s + 5 x})

- \frac{19 s}{( 3 s + 5 x ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{2 s - 3 x}{3 s + 5 x})

Treat

s

as a constant

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

\frac{( - 3 ) ( 3 s + 5 x ) - 5 ( 2 s - 3 x )}{( 3 s + 5 x ) ^{2}} : - \frac{19 s}{( 3 s + 5 x ) ^{2}}

= - \frac{19 s}{( 3 s + 5 x ) ^{2}}

x \to - 2 lim (8 + x^{2})

n = 1 \sum \infty (- \frac{1}{2})^{n}

y^{^{''}} - y^{^{'}} - 2 y = 8, y (0) = 0, y^{^{'}} (0) = 5

x \to 3 lim (4^{x^{3}} - 8 x)

x \to π - lim (sin (x))

What is (\partial)/(\partial x)((2s-3x)/(3s+5x)) ?
The answer to (\partial)/(\partial x)((2s-3x)/(3s+5x)) is -(19s)/((3s+5x)^2)