What is the derivative of f(x)=e^{x^3ln(x)} ?

The derivative of f(x)=e^{x^3ln(x)} is x^{x^3}(3x^2ln(x)+x^2)

What is the first derivative of f(x)=e^{x^3ln(x)} ?

The first derivative of f(x)=e^{x^3ln(x)} is x^{x^3}(3x^2ln(x)+x^2)

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derivative of

f (x) = e^{x^{3} l n (x)}

x^{x^{3}} (3 x^{2} ln (x) + x^{2})

Solution steps

\frac{d}{d x} (e^{x^{3} l n (x)})

Simplify

e^{x^{3} l n (x)} : x^{x^{3}}

= \frac{d}{d x} (x^{x^{3}})

Apply exponent rule:

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

x^{x^{3}} = e^{x^{3} l n (x)}

= \frac{d}{d x} (e^{x^{3} l n (x)})

Apply the chain rule:

e^{x^{3} l n (x)} \frac{d}{d x} (x^{3} ln (x))

= e^{x^{3} l n (x)} \frac{d}{d x} (x^{3} ln (x))

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

= e^{x^{3} l n (x)} (3 x^{2} ln (x) + x^{2})

e^{x^{3} l n (x)} = x^{x^{3}}

= x^{x^{3}} (3 x^{2} ln (x) + x^{2})

\int \frac{9 sin ( x ) + 4 sec ( x )}{tan ( x )} d x

f (x) = \frac{6}{3 x + 1}

f (x) = 8 - 2 x

\int \frac{x ^{3}}{x ^{4} + 1} d x

\frac{\partial}{\partial y} (e^{4 x y})

What is the derivative of f(x)=e^{x^3ln(x)} ?
The derivative of f(x)=e^{x^3ln(x)} is x^{x^3}(3x^2ln(x)+x^2)
What is the first derivative of f(x)=e^{x^3ln(x)} ?
The first derivative of f(x)=e^{x^3ln(x)} is x^{x^3}(3x^2ln(x)+x^2)