What is the derivative of y=x^{4^x} ?

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

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

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

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derivative of

y = x^{4^{x}}

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

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

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

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

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

\int 9 x e^{5 x^{2}} d x

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

x^{2} + x y + 2 y^{2} = 4, (- 2, 1)

9 y^{^{''}} + 6 y^{^{'}} + y = 0, y (- 3) = - 2, y^{^{'}} (1) = - 1

\frac{d}{d x} (x^{2} - x y)

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