What is (\partial)/(\partial y)(yx^{y-1}) ?

The answer to (\partial)/(\partial y)(yx^{y-1}) is x^{y-1}+yx^{y-1}ln(x)

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(\partial)/(\partial y)(yx^{y-1})

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

x^{y - 1} + y x^{y - 1} ln (x)

Solution steps

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

Treat

x

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = x^{y - 1}

= \frac{\partial}{\partial y} (y) x^{y - 1} + \frac{\partial}{\partial y} (x^{y - 1}) y

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

\frac{\partial}{\partial y} (x^{y - 1}) = x^{y - 1} ln (x)

= 1 \cdot x^{y - 1} + x^{y - 1} ln (x) y

Simplify

= x^{y - 1} + y x^{y - 1} ln (x)

y^{^{''}} + 6 y^{^{'}} + 58 y = 0

\frac{d}{d x} (\frac{- x ^{2}}{8})

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

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

\int \frac{11 x ^{2}}{( 64 + x ^{2} ) ^{2}} d x

What is (\partial)/(\partial y)(yx^{y-1}) ?
The answer to (\partial)/(\partial y)(yx^{y-1}) is x^{y-1}+yx^{y-1}ln(x)