What is (\partial)/(\partial y)(7ln(2x^5-3y)) ?

The answer to (\partial)/(\partial y)(7ln(2x^5-3y)) is -(21)/(2x^5-3y)

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(\partial)/(\partial y)(7ln(2x^5-3y))

\frac{\partial}{\partial y} (7 ln (2 x^{5} - 3 y))

- \frac{21}{2 x ^{5} - 3 y}

Solution steps

\frac{\partial}{\partial y} (7 ln (2 x^{5} - 3 y))

Treat

x

as a constant

Take the constant out:

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

= 7 \frac{\partial}{\partial y} (ln (2 x^{5} - 3 y))

Apply the chain rule:

\frac{1}{2 x ^{5} - 3 y} \frac{\partial}{\partial y} (2 x^{5} - 3 y)

= \frac{1}{2 x ^{5} - 3 y} \frac{\partial}{\partial y} (2 x^{5} - 3 y)

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

= 7 \cdot \frac{1}{2 x ^{5} - 3 y} (- 3)

Simplify

7 \cdot \frac{1}{2 x ^{5} - 3 y} (- 3) : - \frac{21}{2 x ^{5} - 3 y}

= - \frac{21}{2 x ^{5} - 3 y}

\frac{1}{10 x ^{2}}, x, \frac{x}{10}

\frac{d y}{d x} = k y

\frac{\partial}{\partial x} (e^{x} \cdot cos (x))

9 (x^{6} - 2)^{5}

\frac{d}{d x} (\frac{x ^{2} - 1}{x + 1})

What is (\partial)/(\partial y)(7ln(2x^5-3y)) ?
The answer to (\partial)/(\partial y)(7ln(2x^5-3y)) is -(21)/(2x^5-3y)