What is the derivative of e^{x+y}-1 ?

The derivative of e^{x+y}-1 is e^{x+y}((dy)/(dx)+1)

What is the first derivative of e^{x+y}-1 ?

The first derivative of e^{x+y}-1 is e^{x+y}((dy)/(dx)+1)

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

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

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

\frac{d}{d x} (1) = 0

= e^{x + y} (1 + \frac{d y}{d x}) - 0

Simplify

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

\int \frac{sin ^{4} ( x )}{cos ^{6} ( x )} d x

f (x) = 7 x, at x = 4

n = 5 \sum \infty \frac{n ^{3}}{2 ^{n}}

f^{^{''}} (x) = 2

x \to - \infty lim (x + \frac{1}{x})

What is the derivative of e^{x+y}-1 ?
The derivative of e^{x+y}-1 is e^{x+y}((dy)/(dx)+1)
What is the first derivative of e^{x+y}-1 ?
The first derivative of e^{x+y}-1 is e^{x+y}((dy)/(dx)+1)