Derivative of Sec Square x

Derivative of sec²x is 2sec²xtanx. Sec²x is the square of the trigonometric function secant x, generally written as sec x.

In this article, we will discuss the derivative of sec^2x, various methods to find it including the chain rule and the quotient rule, solved examples, and some practice problems on it.

What is Derivative of Sec²x?

Derivative of sec²x is 2sec²xtanx. Sec²x is a composite function involving an algebraic operation on a trigonometric function. Derivative of a function gives the rate of change in the functional value with respect to the input variable, i.e. x.

In chain rule, if we need to find the derivative of f(g(x)), it is given as f'(g(x)) × g'(x). The chain rule is one of the most fundamental and used concepts in differential calculus. Formula for the derivative of sec²x can be written as follows:

Derivative of sec²x Formula

Derivative of sec2x formula is added below as,

d/dx[sec²x] = 2sec²x.tanx

We can also represnt it as,

(sec²x)’ = 2sec²x.tanx

It can be derived using,

• Chain Rule of Differentiation
• Quotient Rule
• First Principles of Derivatives

Now let’s learn about them in detial.

Proof of Derivative of sec²x

There are two methods to find derivative of sec²x

• Using Chain Rule of Differentiation
• Using Quotient Rule
• Using First Principles of Derivatives

Derivative of sec²x using Chain Rule of Differentiation

Chain Rule of differentiation states that for a composite function f(g(x)),

[f{g(x)}]’ = f'{g(x)} × g'(x)

Therefore applying chain rule to f(x) = sec²x, we get,

⇒ f'(x) = 2secx × (secx)’

⇒ f'(x) = 2secx × (secx.tanx)

⇒ f'(x) = 2sec²x.tanx

Thus, we have derived the derivative of f(x) = sec²x using the chain rule.

Derivative of sec²x Using Quotient Rule

Quotient rule in differentiation states that,

For two functions u and v the differentiation of (u/v) is found as,

(u/v)’ = (vu’ – uv’)/v²

Now f(x) = sec²x can be written as f(x) = 1/cos²x

Applying quotient rule for f(x) = 1/cos²x, we get,

⇒ f'(x) = (cos²x(1)’ – (1)(cos²x)’)/(cos⁴x)

Now, we know that, (cosx)’ = -sinx

⇒ f'(x) = [-2cosx.(-sinx)]/(cos⁴x)

On simplification, we get

⇒ f'(x) = 2sec²x.tanx

Thus, we obtain the same result for derivative of sec²x by quotient rule.

Derivative of sec²x using First Principle of Derivatives

First principle of differentiation state that derivative of a function f(x) is defined as,

f'(x) = lim_h→0 [f(x + h) – f(x)]/[(x + h) – x]

This can also be represented as,

f'(x) = lim_h→0 [f(x + h) – f(x)]/ h

Putting f(x) = sec²x, to find derivative of sec²x, we get,

⇒ f'(x) = lim_h→0 [sec²(x + h) – sec²x]/ h

⇒ f'(x) = lim_h→0 (sec(x+h) + sec(x)).(sec(x+h) – sec(x))/h

⇒ f'(x) = lim_h→0 (sec(x+h) + sec(x)).(1/cos(x+h) – 1/cos(x))/h

⇒ f'(x) = lim_h→0 (sec(x+h) + sec(x)).(cos(x) – cos(x+h))/hcos(x+h)cos(x)

Using, cos(A + B) = cosAcosB – sinAsinB, we get,

⇒ f'(x) = lim_h→0 (sec(x+h) + sec(x)).(cosx – cosxcosh + sinxsinh)/hcos(x+h)cos(x)

⇒ f'(x) = lim_h→0 (sec(x+h) + sec(x)).(cosx(1 – cosh) + sinxsinh)/hcos(x+h)cos(x)

Now, putting lim_h→0(1-cosh)/h = 0 and lim_h→0(sinh)/h = 1, we get,

⇒ f'(x) = lim_h→0(sec(x+h) + sec(x)).(sinx)/cos(x+h)cosx

⇒ f'(x) = (sec(x+0) + sec(x)).(sinx)/cos(x+0)cosx

⇒ f'(x) = (2secxsinx)/cos²x

⇒ f'(x) = 2sec²xtanx

Thus, derivative of sec²x has been derived using first principle of differentiation.

Read More,

• Derivative in Math
• Derivative of Secant x
• Derivative of Trigonometric Function

Examples on Derivative of sec²x

Various examples on derivative of sec²x

Example 1: Find the derivative of f(x) = sec²(x²+9)

Solution:

We have, f(x) = sec²(x²+9)

By applying chain rule,

⇒ f'(x) = 2sec²(x²+9)×tan(x²+9)×(x²+9)’

⇒ f'(x) = 2sec²(x²+9)×tan(x²+9)×(2x)

⇒ f'(x) = 4x.sec²(x²+9).tan(x²+9)

Example 2: Find the derivative of f(x) = x.sec²x

Solution:

We have, f(x) = xsec²x

By applying product rule,

⇒ f'(x) = x(sec²x)’ + (x)’sec²x

⇒ f'(x) = x.2sec²xtanx + sec²x

⇒ f'(x) = sec²x(2xtanx + 1)

Example 3: Find the derivative of f(x) = x/sec²x

Solution:

We have, f(x) = x/sec²x

By applying product rule,

⇒ f'(x) = [(sec²x)(x)’ – (x)(sec²x)’]/[sec²x]²

⇒ f'(x) = [sec²x – x(2sec²xtanx)]/sec⁴x

⇒ f'(x) = [(sec²x)(1-2xtanx)]/sec⁴x

⇒ f'(x) = (1-2xtanx)/sec²x

⇒ f'(x) = cos²x – 2x.sinx.cosx

⇒ f'(x) = cos²x – xsin2x

Practice Problems on Derivative of sec²x

Some problems on derivative of sec²x

Q1: Find the derivative of the function f(x) = sec²(x²+2x+4)

Q2: Find the derivative of the function f(x) = sec²x + tan²x

Q3: Find the value of f'(x), if f(x) = sec²xtanx.

Q4: If y = sec²x – tan²x, then find the value of dy/dx.

Q5: If y = (sec²x)/x, find the value of dy/dx.

Derivative of sec²x FAQs

What is Derivative of a Function?

Derivative of a function gives the rate of change of the functional value with respect to the input variable. It can also gives slope of tangent to curve at any point on it.

What is Derivative of sec²x?

Derivative of sec²x is 2sec²x.tanx

What are Methods to find Derivative of sec²x?

Methods to find derivative of Root x are,

• First Principle of Differentiation
• Chain Rule
• Quotient Rule

What is Derivative of sec x?

Derivative of secx is secx.tanx

What is Derivative of cos²x?

Derivative of cos²x is -2cosx.sinx