What is the derivative of ln(sec^3(x^4-1)) ?

The derivative of ln(sec^3(x^4-1)) is 12x^3tan(x^4-1)

What is the first derivative of ln(sec^3(x^4-1)) ?

The first derivative of ln(sec^3(x^4-1)) is 12x^3tan(x^4-1)

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\frac{d}{d x} (ln (sec^{3} (x^{4} - 1)))

12 x^{3} tan (x^{4} - 1)

Solution steps

\frac{d}{d x} (ln (sec^{3} (x^{4} - 1)))

Apply the chain rule:

\frac{1}{sec ^{3} ( x ^{4} - 1 )} \frac{d}{d x} (sec^{3} (x^{4} - 1))

= \frac{1}{sec ^{3} ( x ^{4} - 1 )} \frac{d}{d x} (sec^{3} (x^{4} - 1))

\frac{d}{d x} (sec^{3} (x^{4} - 1)) = 12 x^{3} sec^{3} (x^{4} - 1) tan (x^{4} - 1)

= \frac{1}{sec ^{3} ( x ^{4} - 1 )} \cdot 12 x^{3} sec^{3} (x^{4} - 1) tan (x^{4} - 1)

Simplify

\frac{1}{sec ^{3} ( x ^{4} - 1 )} \cdot 12 x^{3} sec^{3} (x^{4} - 1) tan (x^{4} - 1) : 12 x^{3} tan (x^{4} - 1)

= 12 x^{3} tan (x^{4} - 1)

x^{3}

n = 0 \sum \infty cos^{2} (n)

\frac{d}{d x} (\frac{x}{a} + \frac{a}{x})

\frac{\partial}{\partial x} (2 x^{\frac{1}{2}})

\frac{d}{d y} (4 y^{3})

What is the derivative of ln(sec^3(x^4-1)) ?
The derivative of ln(sec^3(x^4-1)) is 12x^3tan(x^4-1)
What is the first derivative of ln(sec^3(x^4-1)) ?
The first derivative of ln(sec^3(x^4-1)) is 12x^3tan(x^4-1)