What is the derivative of y=2e^{2e^x+x} ?

The derivative of y=2e^{2e^x+x} is 2e^{2e^x+x}(2e^x+1)

What is the first derivative of y=2e^{2e^x+x} ?

The first derivative of y=2e^{2e^x+x} is 2e^{2e^x+x}(2e^x+1)

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derivative of

y = 2 e^{2 e^{x} + x}

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

Solution steps

\frac{d}{d x} (2 e^{2 e^{x} + x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (e^{2 e^{x} + x})

Apply the chain rule:

e^{2 e^{x} + x} \frac{d}{d x} (2 e^{x} + x)

= e^{2 e^{x} + x} \frac{d}{d x} (2 e^{x} + x)

\frac{d}{d x} (2 e^{x} + x) = 2 e^{x} + 1

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

\int \frac{x ^{3} + 1}{x ( x - 1 ) ^{3}} d x

f (x) = 3 x^{4} + 2 x - 1, (2, 51)

\int (tan (x))^{7} (sec (x))^{2} d x

x \to 0 + lim (\frac{1}{x} \cdot sin (\frac{1}{x}))

\frac{\partial}{\partial x} (4 ln (x y))

What is the derivative of y=2e^{2e^x+x} ?
The derivative of y=2e^{2e^x+x} is 2e^{2e^x+x}(2e^x+1)
What is the first derivative of y=2e^{2e^x+x} ?
The first derivative of y=2e^{2e^x+x} is 2e^{2e^x+x}(2e^x+1)