What is (\partial)/(\partial y)(3x^2(1+ln(y))) ?

The answer to (\partial)/(\partial y)(3x^2(1+ln(y))) is (3x^2)/y

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(\partial)/(\partial y)(3x^2(1+ln(y)))

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

\frac{3 x ^{2}}{y}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (1) = 0

\frac{\partial}{\partial y} (ln (y)) = \frac{1}{y}

= 3 x^{2} (0 + \frac{1}{y})

Simplify

3 x^{2} (0 + \frac{1}{y}) : \frac{3 x ^{2}}{y}

= \frac{3 x ^{2}}{y}

What is (\partial)/(\partial y)(3x^2(1+ln(y))) ?
The answer to (\partial)/(\partial y)(3x^2(1+ln(y))) is (3x^2)/y