What is (\partial)/(\partial x)(e^{e^y(1+e^x)-2}) ?

The answer to (\partial)/(\partial x)(e^{e^y(1+e^x)-2}) is e^{e^y(1+e^x)-2+x+y}

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(\partial)/(\partial x)(e^{e^y(1+e^x)-2})

\frac{\partial}{\partial x} (e^{e^{y} (1 + e^{x}) - 2})

e^{e^{y} (1 + e^{x}) - 2 + x + y}

Solution steps

\frac{\partial}{\partial x} (e^{e^{y} (1 + e^{x}) - 2})

Treat

y

as a constant

Apply the chain rule:

e^{e^{y} (1 + e^{x}) - 2} \frac{\partial}{\partial x} (e^{y} (1 + e^{x}) - 2)

= e^{e^{y} (1 + e^{x}) - 2} \frac{\partial}{\partial x} (e^{y} (1 + e^{x}) - 2)

\frac{\partial}{\partial x} (e^{y} (1 + e^{x}) - 2) = e^{x + y}

= e^{e^{y} (1 + e^{x}) - 2} e^{x + y}

Apply exponent rule:

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

e^{e^{y} (1 + e^{x}) - 2} e^{x + y} = e^{e^{y} (e^{x} + 1) - 2 + x + y}

= e^{e^{y} (1 + e^{x}) - 2 + x + y}

x^{2} - 2, - (x - 1)^{2} + 3

\int x^{4} (3 x^{5} - 1)^{6} d x

x d y - y d x = x^{2} + y^{2} d x

\frac{d y}{d x} = 5^{x} ln (5) + \frac{1}{x ^{2} + 1}

\frac{d}{d x} (sin (23 x))

What is (\partial)/(\partial x)(e^{e^y(1+e^x)-2}) ?
The answer to (\partial)/(\partial x)(e^{e^y(1+e^x)-2}) is e^{e^y(1+e^x)-2+x+y}