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

The derivative of 4^{x^2} is ln(2)*4^{x^2+1}x

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

The first derivative of 4^{x^2} is ln(2)*4^{x^2+1}x

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derivative of

4^{x^{2}}

ln (2) \cdot 4^{x^{2} + 1} x

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

4^{x^{2}} = e^{x^{2} l n (4)}

= \frac{d}{d x} (e^{x^{2} l n (4)})

Apply the chain rule:

e^{x^{2} l n (4)} \frac{d}{d x} (x^{2} ln (4))

= e^{x^{2} l n (4)} \frac{d}{d x} (x^{2} ln (4))

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

= e^{x^{2} l n (4)} \cdot 4 ln (2) x

Simplify

e^{x^{2} l n (4)} \cdot 4 ln (2) x : ln (2) \cdot 4^{x^{2} + 1} x

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

\int (csc (2 x)) d x

y^{^{'}} - 2 t y = 12 t^{2} e^{t^{2}}, y (0) = 5

x \to 4 lim (\frac{x ^{2} - 4}{x - 2})

\frac{d}{d x} (\frac{3 x + 2}{3 x - 2})

y = x^{4} + 2 x^{2} - x (1.2)

What is the derivative of 4^{x^2} ?
The derivative of 4^{x^2} is ln(2)*4^{x^2+1}x
What is the first derivative of 4^{x^2} ?
The first derivative of 4^{x^2} is ln(2)*4^{x^2+1}x