Question

sec(cot(x³-x+2)),Find the derivative of the given function defined on proper domains.

Answer 1

it is given that function, f(x) = sec(cot(x³ - x + 2))
we have to find the derivative of the given function.

f(x) = sec(cot(x³ - x + 2)))

differentiate with respect to x ,

df(x)/dx = d{sec(cot(x³ - x + 2)))}/dx

= sec(cot(x³ - x + 2)).tan(cot(x³ - x + 2)) × d{cot(x³ - x + 2)}/dx

= sec(cot(x³ - x + 2)). tan(cot(x³ - x + 2)) × {-cosec²(x³ - x + 2))} × d{x³ - x + 2}/dx

= sec(cot(x³ - x + 2)). tan(cot(x³ - x + 2)) × {-cosec²(x³ - x + 2))} × (3x² - 1)

= -(3x² - 1) . sec(cot(x³ - x + 2)). tan(cot(x³ - x + 2)) cosec²(x³ - x + 2)

hence, first order derivative of the given function is -(3x² - 1) . sec(cot(x³ - x + 2)). tan(cot(x³ - x + 2)) cosec²(x³ - x + 2)

Answer 2

HELLO DEAR,

let y = sec{cot(x³ - x + 2)}

put t = x³ - x + 2, v = cot(t) , y = sec(v)

now, dy/dx = dy/dv × dv/dt × dt/dx

therefore, dy/dx = (secv)/dv × d{cot(t)}/dt × d(x³ - x + 2)/dx

=> dy/dx = (secv.tanv) × (-cosec²t) × (3x² - 1)

=> dy/dx = -cosec²t×secv×tanv×(3x² - 1)

put v = t = (x³ - x + 2) , v = cot (x³ - x + 2)

therefore, dy/dx = -cosec²(x³ - x + 2) × sec{cot(x³ - x + 2)} × tan{cot(x³ - x + 2)} × (3x² - 1)

I HOPE ITS HELP YOU DEAR,
THANKS