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

The derivative of 9e^x+x is 9e^x+1

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

The first derivative of 9e^x+x is 9e^x+1

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derivative of

9 e^{x} + x

9 e^{x} + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d x}{d x} = 1

= 9 e^{x} + 1

\frac{d}{d x} (ln (\frac{x - 1}{x + 1}))

(x^{2} + 1) \frac{d y}{d x} + 2 x y - x = 0

x \to - 2 lim (\frac{x ^{2} + 8}{x + 2})

x^{^{''}} + 100 x = 0

\int x cos^{2} (x^{2}) d x

What is the derivative of 9e^x+x ?
The derivative of 9e^x+x is 9e^x+1
What is the first derivative of 9e^x+x ?
The first derivative of 9e^x+x is 9e^x+1