What is the derivative of y=9e^{9e^x+x} ?

The derivative of y=9e^{9e^x+x} is 9e^{9e^x+x}(9e^x+1)

What is the first derivative of y=9e^{9e^x+x} ?

The first derivative of y=9e^{9e^x+x} is 9e^{9e^x+x}(9e^x+1)

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derivative of

y = 9 e^{9 e^{x} + x}

9 e^{9 e^{x} + x} (9 e^{x} + 1)

Solution steps

\frac{d}{d x} (9 e^{9 e^{x} + x})

Take the constant out:

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

= 9 \frac{d}{d x} (e^{9 e^{x} + x})

Apply the chain rule:

e^{9 e^{x} + x} \frac{d}{d x} (9 e^{x} + x)

= e^{9 e^{x} + x} \frac{d}{d x} (9 e^{x} + x)

\frac{d}{d x} (9 e^{x} + x) = 9 e^{x} + 1

= 9 e^{9 e^{x} + x} (9 e^{x} + 1)

1 + e^{x}

y^{^{''}} + 2 y^{^{'}} + y = 8 e^{- t}

\frac{d y}{d x} = - \frac{2 x y + y ^{2} + 1}{x ^{2} + 2 x y}

k = 1 \sum \infty \frac{2 ^{k}}{k !}

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

What is the derivative of y=9e^{9e^x+x} ?
The derivative of y=9e^{9e^x+x} is 9e^{9e^x+x}(9e^x+1)
What is the first derivative of y=9e^{9e^x+x} ?
The first derivative of y=9e^{9e^x+x} is 9e^{9e^x+x}(9e^x+1)