What is (\partial)/(\partial x)(7e^{x^2ln(y)}) ?

The answer to (\partial)/(\partial x)(7e^{x^2ln(y)}) is 14xy^{x^2}ln(y)

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(\partial)/(\partial x)(7e^{x^2ln(y)})

\frac{\partial}{\partial x} (7 e^{x^{2} l n (y)})

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

Solution steps

\frac{\partial}{\partial x} (7 e^{x^{2} l n (y)})

Simplify

7 e^{x^{2} l n (y)} : 7 y^{x^{2}}

= \frac{\partial}{\partial x} (7 y^{x^{2}})

Treat

y

as a constant

Take the constant out:

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

= 7 \frac{\partial}{\partial x} (y^{x^{2}})

Apply exponent rule:

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

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

= 7 \frac{\partial}{\partial x} (e^{x^{2} l n (y)})

Apply the chain rule:

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

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

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

= 7 e^{x^{2} l n (y)} \cdot 2 x ln (y)

Simplify

7 e^{x^{2} l n (y)} \cdot 2 x ln (y) : 14 x y^{x^{2}} ln (y)

= 14 x y^{x^{2}} ln (y)

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

\int \frac{7}{x ^{2} x ^{2} + 9} d x

\int \frac{1}{y ( 1 - y )} d y

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

(2 + x) y^{^{'}} = 8 y

What is (\partial)/(\partial x)(7e^{x^2ln(y)}) ?
The answer to (\partial)/(\partial x)(7e^{x^2ln(y)}) is 14xy^{x^2}ln(y)