What is the derivative of 4e^{4e^x+x} ?

The derivative of 4e^{4e^x+x} is 4e^{4e^x+x}(4e^x+1)

What is the first derivative of 4e^{4e^x+x} ?

The first derivative of 4e^{4e^x+x} is 4e^{4e^x+x}(4e^x+1)

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\frac{d}{d x} (4 e^{4 e^{x} + x})

4 e^{4 e^{x} + x} (4 e^{x} + 1)

Solution steps

\frac{d}{d x} (4 e^{4 e^{x} + x})

Take the constant out:

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

= 4 \frac{d}{d x} (e^{4 e^{x} + x})

Apply the chain rule:

e^{4 e^{x} + x} \frac{d}{d x} (4 e^{x} + x)

= e^{4 e^{x} + x} \frac{d}{d x} (4 e^{x} + x)

\frac{d}{d x} (4 e^{x} + x) = 4 e^{x} + 1

= 4 e^{4 e^{x} + x} (4 e^{x} + 1)

f (x) = \frac{- x ^{2} - 4}{( x ^{2} - 4 ) ^{2}}

\frac{d ^{2}}{d x ^{2}} (x^{3})

x^{2} arctan (x)

\int 3 sin^{3} (x co) s^{5} x d x

\int \frac{1}{( x ^{2} + 4 ) ^{3}} d x

What is the derivative of 4e^{4e^x+x} ?
The derivative of 4e^{4e^x+x} is 4e^{4e^x+x}(4e^x+1)
What is the first derivative of 4e^{4e^x+x} ?
The first derivative of 4e^{4e^x+x} is 4e^{4e^x+x}(4e^x+1)