What is (\partial)/(\partial x)(1/(y^2+ln(x))) ?

The answer to (\partial)/(\partial x)(1/(y^2+ln(x))) is -1/(x(y^2+ln(x))^2)

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(\partial)/(\partial x)(1/(y^2+ln(x)))

\frac{\partial}{\partial x} (\frac{1}{y ^{2} + ln ( x )})

- \frac{1}{x ( y ^{2} + ln ( x ) ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{1}{y ^{2} + ln ( x )})

Treat

y

as a constant

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (y^{2} + ln (x)) = \frac{1}{x}

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

Simplify

- \frac{1}{( y ^{2} + ln ( x ) ) ^{2}} \cdot \frac{1}{x} : - \frac{1}{x ( y ^{2} + ln ( x ) ) ^{2}}

= - \frac{1}{x ( y ^{2} + ln ( x ) ) ^{2}}

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What is (\partial)/(\partial x)(1/(y^2+ln(x))) ?
The answer to (\partial)/(\partial x)(1/(y^2+ln(x))) is -1/(x(y^2+ln(x))^2)