Question

find derivative of sec(cot(x^3-x+2))​

Answer 1

Step-by-step explanation:

By using the quotient rule and trigonometric identities, we can obtain the following derivatives:

\displaystyle\frac{{{d}{\left( \csc{{x}}\right)}}}{{{\left.{d}{x}\right.}}}=- \csc{{x}} \cot{{x}}

dx

d(cscx)

=−cscxcotx

\displaystyle\frac{{{d}{\left( \sec{{x}}\right)}}}{{{\left.{d}{x}\right.}}}= \sec{{x}} \tan{{x}}

dx

d(secx)

=secxtanx

\displaystyle\frac{{{d}{\left( \cot{{x}}\right)}}}{{{\left.{d}{x}\right.}}}=-{{\csc}^{2}{x}}

dx

d(cotx)

=−csc

2

x

In words, we would say:

The derivative of \displaystyle \csc{{x}}cscx is \displaystyle- \csc{{x}} \cot{{x}}−cscxcotx,

The derivative of \displaystyle \sec{{x}}secx is \displaystyle \sec{{x}} \tan{{x}}secxtanx and

The derivative of \displaystyle \cot{{x}}cotx is \displaystyle-{{\csc}^{2}{x}}−csc

2

x.