What is (\partial)/(\partial x)(y^{x^2-y}) ?

The answer to (\partial)/(\partial x)(y^{x^2-y}) is 2xy^{x^2-y}ln(y)

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(\partial)/(\partial x)(y^{x^2-y})

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

2 x y^{x^{2} - y} ln (y)

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

a^{b} = e^{b l n (a)}

y^{x^{2} - y} = e^{(x^{2} - y) l n (y)}

= \frac{\partial}{\partial x} (e^{(x^{2} - y) l n (y)})

Apply the chain rule:

e^{(x^{2} - y) l n (y)} \frac{\partial}{\partial x} ((x^{2} - y) ln (y))

= e^{(x^{2} - y) l n (y)} \frac{\partial}{\partial x} ((x^{2} - y) ln (y))

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

= e^{(x^{2} - y) l n (y)} \cdot 2 x ln (y)

e^{(x^{2} - y) l n (y)} = y^{x^{2} - y}

= 2 x y^{x^{2} - y} ln (y)

3 t - 4 t^{2}

\frac{d y}{d x} y = cos^{- 1} (\frac{1}{x})

(3.4) (- 6.4)

x \to 0 - lim (ln (x^{8}))

\frac{8}{x ^{2} + 4}

What is (\partial)/(\partial x)(y^{x^2-y}) ?
The answer to (\partial)/(\partial x)(y^{x^2-y}) is 2xy^{x^2-y}ln(y)