What is (\partial)/(\partial x)(a/(ln(x)-c)) ?

The answer to (\partial)/(\partial x)(a/(ln(x)-c)) is -a/(x(ln(x)-c)^2)

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(\partial)/(\partial x)(a/(ln(x)-c))

\frac{\partial}{\partial x} (\frac{a}{ln ( x ) - c})

- \frac{a}{x ( ln ( x ) - c ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{a}{ln ( x ) - c})

Take the constant out:

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

= a \frac{\partial}{\partial x} (\frac{1}{ln ( x ) - c})

Apply exponent rule:

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

= a \frac{\partial}{\partial x} ((ln (x) - c)^{- 1})

Apply the chain rule:

- \frac{1}{( ln ( x ) - c ) ^{2}} \frac{\partial}{\partial x} (ln (x) - c)

= - \frac{1}{( ln ( x ) - c ) ^{2}} \frac{\partial}{\partial x} (ln (x) - c)

\frac{\partial}{\partial x} (ln (x) - c) = \frac{1}{x}

= a (- \frac{1}{( ln ( x ) - c ) ^{2}} \cdot \frac{1}{x})

Simplify

a (- \frac{1}{( ln ( x ) - c ) ^{2}} \cdot \frac{1}{x}) : - \frac{a}{x ( ln ( x ) - c ) ^{2}}

= - \frac{a}{x ( ln ( x ) - c ) ^{2}}

y^{^{'}} + \frac{4 y}{x} = - \frac{4}{x}

f (x) = \frac{6 x}{x ^{2} - 3}, at x = 3

\frac{\partial}{\partial z} (y e^{x z})

9 y^{^{''}} - 6 y^{^{'}} + 8 y = 0

\frac{d y}{d x} = 3 x^{2} (y - 1)^{3}

What is (\partial)/(\partial x)(a/(ln(x)-c)) ?
The answer to (\partial)/(\partial x)(a/(ln(x)-c)) is -a/(x(ln(x)-c)^2)