What is the derivative of 2^{4x}+4x^2 ?

The derivative of 2^{4x}+4x^2 is ln(2)*2^{4x+2}+8x

What is the first derivative of 2^{4x}+4x^2 ?

The first derivative of 2^{4x}+4x^2 is ln(2)*2^{4x+2}+8x

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derivative of

2^{4 x} + 4 x^{2}

ln (2) \cdot 2^{4 x + 2} + 8 x

Solution steps

\frac{d}{d x} (2^{4 x} + 4 x^{2})

Simplify

2^{4 x} + 4 x^{2} : 1 6^{x} + 4 x^{2}

= \frac{d}{d x} (1 6^{x} + 4 x^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (1 6^{x}) + \frac{d}{d x} (4 x^{2})

\frac{d}{d x} (1 6^{x}) = ln (2) \cdot 2^{4 x + 2}

\frac{d}{d x} (4 x^{2}) = 8 x

= ln (2) \cdot 2^{4 x + 2} + 8 x

\int \frac{1}{4 ^{2 x}} d x

ln (ln (4 s))

\frac{\partial}{\partial x} (x^{3} + 2 x y + y^{2})

\frac{d}{d x} (0.015 + x)

f (x) = 6 x^{2} - 10 x + 4, at x = 1

What is the derivative of 2^{4x}+4x^2 ?
The derivative of 2^{4x}+4x^2 is ln(2)*2^{4x+2}+8x
What is the first derivative of 2^{4x}+4x^2 ?
The first derivative of 2^{4x}+4x^2 is ln(2)*2^{4x+2}+8x