from a handpicked tutor in LIVE 1-to-1 classes

# Differentiate the function with respect to x. sec(tan(√x))

**Solution:**

Let f(x) = sec (tan (√x)),

u(x) = √x ,

v(t) = tan t and w(s) = sec s

Then, (w o v o u) (x) = w[ v ( u ( x ) ) ]

= w[v ( √ x )] = w(tan √x)

= sec (tan √x) = f(x)

Here, f is a composite function of three functions u, v, and w.

Put, s = v(t) = tant

and t = u(x) = √x

Then,

⇒ dw / ds = d / ds(sec s)

= sec s tan s = sec (tan t) .tan (tan t)

[s = tant]

= sec (tan √x) .tan (tan √x)

[t = √x]

Now,

ds/dt = d/dt(tan t) = sec 2t = sec 2√x dt/dx

= d / dx (√x) = d / dx (x^{1/2})

= 1/2.x^{12−1 }

= 1/2√x

Hence, by chain rule, we get

d/dx [sec (tan √x)]

= dw / ds. ds / dt .dt / dx

= sec (tan √x) .tan (tan √x) .sec^{2} √x. 1/2 √x

= 1/2 √x sec^{2}√x sec (tan √x) tan (tan √x)

= sec^{2} √x sec (tan√x) tan (tan √x) / 2√x

Alternate method:

d / dx [sec (tan √x)] = sec (tan √x). tan (tan √x). d / dx (tan √x)

= sec (tan √x).tan (tan √x).sec^{2}(√x) .d / dx (√x)

= sec (tan √x).tan (tan √x).sec^{2} (√x).1/2 √x

= sec (tan √x).tan (tan √x) .sec^{2} (√x) / 2√x

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.2 Question 4

## Differentiate the function with respect to x. sec(tan(√x))

**Summary:**

By chain rule we have obtained the derivative of sec(tan(√x)) with respect to x is sec (tan √x).tan (tan √x) .sec^{2} (√x) / 2√x

visual curriculum