What is the d/(dy)(e^{xe^y}) ?

The d/(dy)(e^{xe^y}) is e^{e^yx+y}x

What is the first d/(dy)(e^{xe^y}) ?

The first d/(dy)(e^{xe^y}) is e^{e^yx+y}x

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

e^{e^{y} x + y} x

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

= e^{x e^{y}} x e^{y}

Apply exponent rule:

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

e^{x e^{y}} e^{y} = e^{e^{y} x + y}

= e^{e^{y} x + y} x

x, x - 2, y = 0

x \to 0 lim (\frac{sin ( k x )}{x})

\int \frac{3 r}{1 - r ^{2}} d r

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

y = (\frac{x}{3.3} + \frac{3.3}{x}) (x^{2} + 1)