What is the derivative of y=x^3+e^x ?

The derivative of y=x^3+e^x is 3x^2+e^x

What is the first derivative of y=x^3+e^x ?

The first derivative of y=x^3+e^x is 3x^2+e^x

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derivative of

y = x^{3} + e^{x}

3 x^{2} + e^{x}

Solution steps

\frac{d}{d x} (x^{3} + e^{x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{3}) + \frac{d}{d x} (e^{x})

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

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

= 3 x^{2} + e^{x}

y^{^{''}} - 4 y^{^{'}} + 5 y = 1 + 5 x

\int 9 x + 6 d x

\int \frac{x}{x ^{4} + 3 x ^{2} - 4} d x

n = 1 \sum \infty \frac{n + 2}{n !}

\frac{d N}{d t} = N (1 - 0.0004 N)

What is the derivative of y=x^3+e^x ?
The derivative of y=x^3+e^x is 3x^2+e^x
What is the first derivative of y=x^3+e^x ?
The first derivative of y=x^3+e^x is 3x^2+e^x