What is (\partial)/(\partial y)(e^xln(1+y)) ?

The answer to (\partial)/(\partial y)(e^xln(1+y)) is (e^x)/(1+y)

Popular >

(\partial)/(\partial y)(e^xln(1+y))

\frac{\partial}{\partial y} (e^{x} ln (1 + y))

\frac{e ^{x}}{1 + y}

Solution steps

\frac{\partial}{\partial y} (e^{x} ln (1 + y))

Treat

x

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= e^{x} \frac{1}{1 + y} \cdot 1

Simplify

e^{x} \frac{1}{1 + y} \cdot 1 : \frac{e ^{x}}{1 + y}

= \frac{e ^{x}}{1 + y}

f (x) = x sin (x)

y^{^{''}} + \frac{1}{4} y^{^{'}} + 2 y = 2 cos (4123 t)

\int \frac{1 - 4 x ^{2}}{x} d x

\frac{d}{d x} (\frac{2 x ^{2} + 2 x + 6}{x})

\frac{2}{3 t ^{2}}

What is (\partial)/(\partial y)(e^xln(1+y)) ?
The answer to (\partial)/(\partial y)(e^xln(1+y)) is (e^x)/(1+y)