What is the d/(dy)(x^{(1+y)}) ?

The d/(dy)(x^{(1+y)}) is x^{1+y}ln(x)

What is the first d/(dy)(x^{(1+y)}) ?

The first d/(dy)(x^{(1+y)}) is x^{1+y}ln(x)

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\frac{d}{d y} (x^{(1 + y)})

x^{1 + y} ln (x)

Solution steps

\frac{d}{d y} (x^{(1 + y)})

Treat

x

as a constant

Apply exponent rule:

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

x^{1 + y} = e^{(1 + y) l n (x)}

= \frac{d}{d y} (e^{(1 + y) l n (x)})

Apply the chain rule:

e^{(1 + y) l n (x)} \frac{d}{d y} ((1 + y) ln (x))

= e^{(1 + y) l n (x)} \frac{d}{d y} ((1 + y) ln (x))

\frac{d}{d y} ((1 + y) ln (x)) = ln (x)

= e^{(1 + y) l n (x)} ln (x)

e^{(1 + y) l n (x)} = x^{1 + y}

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

\frac{d}{d x} (\frac{d y}{d x}) + y = 0

x \to - 1.8 lim (- 3 x + 18.4)

y^{^{'}} = e^{t + y}

\frac{\partial}{\partial x} (- 3 cos (x))

\frac{1}{s ( 12 s ^{2} + 10 s + 1 )}

What is the d/(dy)(x^{(1+y)}) ?
The d/(dy)(x^{(1+y)}) is x^{1+y}ln(x)
What is the first d/(dy)(x^{(1+y)}) ?
The first d/(dy)(x^{(1+y)}) is x^{1+y}ln(x)