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

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

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

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

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derivative of

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)

\frac{d}{d x} (6 x + 6)

y^{^{''}} + 9 y = sec (3 x)

\int \frac{6 z}{( 5 - 3 z ^{2} ) ^{2}} d z

2 \frac{d y}{d x} = y^{2}

\int 7 x e^{x^{2}} d x

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