What is the derivative of (e^{3x}+1^2) ?

The derivative of (e^{3x}+1^2) is 6e^{3x}(e^{3x}+1)

What is the first derivative of (e^{3x}+1^2) ?

The first derivative of (e^{3x}+1^2) is 6e^{3x}(e^{3x}+1)

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

6 e^{3 x} (e^{3 x} + 1)

Solution steps

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

Apply the chain rule:

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

= 2 (e^{3 x} + 1) \frac{d}{d x} (e^{3 x} + 1)

\frac{d}{d x} (e^{3 x} + 1) = 3 e^{3 x}

= 2 (e^{3 x} + 1) \cdot 3 e^{3 x}

Simplify

= 6 e^{3 x} (e^{3 x} + 1)

3 t sec (t)

t \cdot sin (2 t)

y^{^{''}} - y^{^{'}} + y = 13 sin (t)

x \to 0 lim (\frac{4 x}{sin ( x )})

f (x) = 2 + \frac{7}{x} + \frac{2}{x ^{2}}

What is the derivative of (e^{3x}+1^2) ?
The derivative of (e^{3x}+1^2) is 6e^{3x}(e^{3x}+1)
What is the first derivative of (e^{3x}+1^2) ?
The first derivative of (e^{3x}+1^2) is 6e^{3x}(e^{3x}+1)