Use the quotient rule to differentiate the function f(x) = tan(x) - 1/ sec(x) .

Solution:

Given, f(x) = [tan(x) - 1]/sec(x)

We have to differentiate f(x) using the quotient rule.

The Quotient rule is given by

f(x) = g(x)/h(x)

f’(x) = {[g’(x)h(x)] - [g(x)h’(x)]}/h(x)²

Here, g(x) = tan(x) - 1

g’(x) = sec²(x)

h(x) = sec(x)

h’(x) = sec(x) tan(x)

f’(x) = \(\frac{(sec^{2}(x)\times sec(x))-((tan(x)-1)\times sec(x)tan(x))}{sec^{2}(x)}\)

\(=\frac{sec^{3}(x)-sec(x)tan^{2}(x)+sec(x)tan(x)}{sec^{2}(x)}\\=sec(x)+\frac{tan(x)-tan^{2}(x)}{sec(x)}\)

We know, tan(x) = sin(x)/cos(x)

sec(x) = 1/cos(x)

So, tan(x)/sec(x) = [sin(x)/cos(x)]/[1/cos(x)] = sin(x)

f’(x) = sec(x) + sin(x) - sin(x)tan(x)

f’(x) = \(\frac{1}{cos(x)}+sin(x)-\frac{sin(x)sin(x)}{cos(x)}\\f'(x)=\frac{1}{cos(x)}+sin(x)-\frac{sin^{2}(x)}{cos(x)}\\f'(x)=\frac{1+sin(x)cos(x)-sin^{2}(x)}{cos(x)}\)

We know, \(sin^{2}(x)+cos^{2}(x)=1\)

\(f'(x)=\frac{cos^{2}(x)+sin(x)cos(x)}{cos(x)}\)

f’(x) = cos(x) + sin(x)

Therefore, the derivative of f(x) is cos(x) + sin(x).

---

Use the quotient rule to differentiate the function f(x) = tan(x) - 1/ sec(x) .

Summary:

Using the quotient rule to differentiate the function f(x) = tan(x) - 1/ sec(x) is cos(x) + sin(x).