What is the derivative of f(x)=(5+x^4)^{3x} ?

The derivative of f(x)=(5+x^4)^{3x} is 3(ln(5+x^4)+(4x^4)/(5+x^4))(5+x^4)^{3x}

What is the first derivative of f(x)=(5+x^4)^{3x} ?

The first derivative of f(x)=(5+x^4)^{3x} is 3(ln(5+x^4)+(4x^4)/(5+x^4))(5+x^4)^{3x}

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derivative of

f (x) = (5 + x^{4})^{3 x}

3 (ln (5 + x^{4}) + \frac{4 x ^{4}}{5 + x ^{4}}) (5 + x^{4})^{3 x}

Solution steps

\frac{d}{d x} ((5 + x^{4})^{3 x})

Apply exponent rule:

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

(5 + x^{4})^{3 x} = e^{3 x l n (5 + x^{4})}

= \frac{d}{d x} (e^{3 x l n (5 + x^{4})})

Apply the chain rule:

e^{3 x l n (5 + x^{4})} \frac{d}{d x} (3 x ln (5 + x^{4}))

= e^{3 x l n (5 + x^{4})} \frac{d}{d x} (3 x ln (5 + x^{4}))

\frac{d}{d x} (3 x ln (5 + x^{4})) = 3 (ln (5 + x^{4}) + \frac{4 x ^{4}}{5 + x ^{4}})

= e^{3 x l n (5 + x^{4})} \cdot 3 (ln (5 + x^{4}) + \frac{4 x ^{4}}{5 + x ^{4}})

e^{3 x l n (5 + x^{4})} = (5 + x^{4})^{3 x}

= 3 (ln (5 + x^{4}) + \frac{4 x ^{4}}{5 + x ^{4}}) (5 + x^{4})^{3 x}

\frac{\partial}{\partial y} (x^{3} + k y^{2} - 5 x y)

\frac{d}{d x} ((x^{2} - 7)^{2})

y = - \frac{8 x}{x ^{2} + 1}

f (x) = \frac{e ^{x}}{x + 6}, (0, \frac{1}{6})

f (x) = (x + 2 x) e^{x}

What is the derivative of f(x)=(5+x^4)^{3x} ?
The derivative of f(x)=(5+x^4)^{3x} is 3(ln(5+x^4)+(4x^4)/(5+x^4))(5+x^4)^{3x}
What is the first derivative of f(x)=(5+x^4)^{3x} ?
The first derivative of f(x)=(5+x^4)^{3x} is 3(ln(5+x^4)+(4x^4)/(5+x^4))(5+x^4)^{3x}