What is (\partial)/(\partial y)(x^2ln(3y-5)) ?

The answer to (\partial)/(\partial y)(x^2ln(3y-5)) is (3x^2)/(3y-5)

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(\partial)/(\partial y)(x^2ln(3y-5))

\frac{\partial}{\partial y} (x^{2} ln (3 y - 5))

\frac{3 x ^{2}}{3 y - 5}

Solution steps

\frac{\partial}{\partial y} (x^{2} ln (3 y - 5))

Treat

x

as a constant

Take the constant out:

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

= x^{2} \frac{\partial}{\partial y} (ln (3 y - 5))

Apply the chain rule:

\frac{1}{3 y - 5} \frac{\partial}{\partial y} (3 y - 5)

= \frac{1}{3 y - 5} \frac{\partial}{\partial y} (3 y - 5)

\frac{\partial}{\partial y} (3 y - 5) = 3

= x^{2} \frac{1}{3 y - 5} \cdot 3

Simplify

x^{2} \frac{1}{3 y - 5} \cdot 3 : \frac{3 x ^{2}}{3 y - 5}

= \frac{3 x ^{2}}{3 y - 5}

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

n = 2 \sum \infty \frac{1}{n ^{2} - 1}

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

x \to 3 lim (\frac{x ^{2} + x}{x - 3})

\int_{0}^{k} \frac{1}{2} (e^{x} - 1) d x

What is (\partial)/(\partial y)(x^2ln(3y-5)) ?
The answer to (\partial)/(\partial y)(x^2ln(3y-5)) is (3x^2)/(3y-5)