What is (\partial)/(\partial x)(y*ln(x^2+5y+40)) ?

The answer to (\partial)/(\partial x)(y*ln(x^2+5y+40)) is (2yx)/(x^2+5y+40)

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(\partial)/(\partial x)(y*ln(x^2+5y+40))

\frac{\partial}{\partial x} (y \cdot ln (x^{2} + 5 y + 40))

\frac{2 y x}{x ^{2} + 5 y + 40}

Solution steps

\frac{\partial}{\partial x} (y ln (x^{2} + 5 y + 40))

Treat

y

as a constant

Take the constant out:

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

= y \frac{\partial}{\partial x} (ln (x^{2} + 5 y + 40))

Apply the chain rule:

\frac{1}{x ^{2} + 5 y + 40} \frac{\partial}{\partial x} (x^{2} + 5 y + 40)

= \frac{1}{x ^{2} + 5 y + 40} \frac{\partial}{\partial x} (x^{2} + 5 y + 40)

\frac{\partial}{\partial x} (x^{2} + 5 y + 40) = 2 x

= y \frac{1}{x ^{2} + 5 y + 40} \cdot 2 x

Simplify

y \frac{1}{x ^{2} + 5 y + 40} \cdot 2 x : \frac{2 y x}{x ^{2} + 5 y + 40}

= \frac{2 y x}{x ^{2} + 5 y + 40}

\frac{d}{d t} (e^{\frac{t}{2}})

x \to \infty + lim (5 x + 6)

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

x \to 2 - lim (1 - x^{2})

\int_{1}^{t} \frac{29}{x ^{3}} d x

What is (\partial)/(\partial x)(y*ln(x^2+5y+40)) ?
The answer to (\partial)/(\partial x)(y*ln(x^2+5y+40)) is (2yx)/(x^2+5y+40)