What is (\partial)/(\partial x)(ln(x^6+1)+y^2) ?

The answer to (\partial)/(\partial x)(ln(x^6+1)+y^2) is (6x^5)/(x^6+1)

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(\partial)/(\partial x)(ln(x^6+1)+y^2)

\frac{\partial}{\partial x} (ln (x^{6} + 1) + y^{2})

\frac{6 x ^{5}}{x ^{6} + 1}

Solution steps

\frac{\partial}{\partial x} (ln (x^{6} + 1) + y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (ln (x^{6} + 1)) + \frac{\partial}{\partial x} (y^{2})

\frac{\partial}{\partial x} (ln (x^{6} + 1)) = \frac{6 x ^{5}}{x ^{6} + 1}

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

= \frac{6 x ^{5}}{x ^{6} + 1} + 0

Simplify

= \frac{6 x ^{5}}{x ^{6} + 1}

\frac{d}{d x} (\frac{5}{12} (3 + 5) x^{3})

x \to - \infty lim (x^{0})

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

\frac{d}{d x} ((4 x^{2} - 3 x) e^{5 x})

\int (cos^{2} (x))^{2} d x

What is (\partial)/(\partial x)(ln(x^6+1)+y^2) ?
The answer to (\partial)/(\partial x)(ln(x^6+1)+y^2) is (6x^5)/(x^6+1)