What is (\partial)/(\partial x)(ln(1+x^2)-2e^{5y}) ?

The answer to (\partial)/(\partial x)(ln(1+x^2)-2e^{5y}) is (2x)/(x^2+1)

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(\partial)/(\partial x)(ln(1+x^2)-2e^{5y})

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

\frac{2 x}{x ^{2} + 1}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{\partial}{\partial x} (2 e^{5 y}) = 0

= \frac{2 x}{1 + x ^{2}} - 0

Simplify

= \frac{2 x}{x ^{2} + 1}

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\frac{d}{d x} (\frac{e ^{x}}{5 x})

\int - 3 arcsin (10 x) d x

What is (\partial)/(\partial x)(ln(1+x^2)-2e^{5y}) ?
The answer to (\partial)/(\partial x)(ln(1+x^2)-2e^{5y}) is (2x)/(x^2+1)