What is the d/(dy)(e^yy^e) ?

The d/(dy)(e^yy^e) is e^yy^e+e^{1+y}y^{e-1}

What is the first d/(dy)(e^yy^e) ?

The first d/(dy)(e^yy^e) is e^yy^e+e^{1+y}y^{e-1}

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\frac{d}{d y} (e^{y} y^{e})

e^{y} y^{e} + e^{1 + y} y^{e - 1}

Solution steps

\frac{d}{d y} (e^{y} y^{e})

Apply the Product Rule:

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

f = e^{y}, g = y^{e}

= \frac{d}{d y} (e^{y}) y^{e} + \frac{d}{d y} (y^{e}) e^{y}

\frac{d}{d y} (e^{y}) = e^{y}

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

= e^{y} y^{e} + e y^{e - 1} e^{y}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e e^{y} = e^{1 + y}

= e^{y} y^{e} + e^{1 + y} y^{e - 1}

\int 4 e^{x} cos (2 x) d x

f (x) = (1 + 3 x) (5 - 2 x), at x = 2

\frac{s}{( s + 1 ) ( s ^{2} + 4 )}

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x ^{10}})

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What is the d/(dy)(e^yy^e) ?
The d/(dy)(e^yy^e) is e^yy^e+e^{1+y}y^{e-1}
What is the first d/(dy)(e^yy^e) ?
The first d/(dy)(e^yy^e) is e^yy^e+e^{1+y}y^{e-1}