What is the derivative of f(x)=2^{4x}+4x^2 ?

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

What is the first derivative of f(x)=2^{4x}+4x^2 ?

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

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derivative of

f (x) = 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

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

\frac{d y}{d x}, 5 x y = 8

x \to 0 lim (\frac{x}{x + 7})

y^{^{'}} = 1 + y - 2 x + 3

\int (x - 1)^{2} (x + 1)^{2} d x

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