What is (\partial)/(\partial x)(9ln(8x^2+y^3)) ?

The answer to (\partial)/(\partial x)(9ln(8x^2+y^3)) is (144x)/(8x^2+y^3)

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(\partial)/(\partial x)(9ln(8x^2+y^3))

\frac{\partial}{\partial x} (9 ln (8 x^{2} + y^{3}))

\frac{144 x}{8 x ^{2} + y ^{3}}

Solution steps

\frac{\partial}{\partial x} (9 ln (8 x^{2} + y^{3}))

Treat

y

as a constant

Take the constant out:

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

= 9 \frac{\partial}{\partial x} (ln (8 x^{2} + y^{3}))

Apply the chain rule:

\frac{1}{8 x ^{2} + y ^{3}} \frac{\partial}{\partial x} (8 x^{2} + y^{3})

= \frac{1}{8 x ^{2} + y ^{3}} \frac{\partial}{\partial x} (8 x^{2} + y^{3})

\frac{\partial}{\partial x} (8 x^{2} + y^{3}) = 16 x

= 9 \cdot \frac{1}{8 x ^{2} + y ^{3}} \cdot 16 x

Simplify

9 \cdot \frac{1}{8 x ^{2} + y ^{3}} \cdot 16 x : \frac{144 x}{8 x ^{2} + y ^{3}}

= \frac{144 x}{8 x ^{2} + y ^{3}}

\int e^{8 x} sin (x) d x

\frac{\partial}{\partial x} (x^{2} sin (\frac{1}{x}))

\frac{1}{x ( x + 1 )}

\int \frac{3 x ^{2} + 4 x + 1}{3 x ^{3} + 6 x ^{2} + 3 x + 5} d x

What is (\partial)/(\partial x)(9ln(8x^2+y^3)) ?
The answer to (\partial)/(\partial x)(9ln(8x^2+y^3)) is (144x)/(8x^2+y^3)