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
Also Check, Trigonometric Function
It can be derived using,
- Chain Rule of Differentiation
- Quotient Rule
- First Principles of Derivatives
Now let’s learn about them in detail.
Read: Calculus in Maths
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) = limh→0 [f(x + h) – f(x)]/[(x + h) – x]
This can also be represented as,
f'(x) = limh→0 [f(x + h) – f(x)]/ h
Putting f(x) = sec2x, to find derivative of sec2x, we get,
⇒ f'(x) = limh→0 [sec2(x + h) – sec2x]/ h
⇒ f'(x) = limh→0 (sec(x+h) + sec(x)).(sec(x+h) – sec(x))/h
⇒ f'(x) = limh→0 (sec(x+h) + sec(x)).(1/cos(x+h) – 1/cos(x))/h
⇒ f'(x) = limh→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) = limh→0 (sec(x+h) + sec(x)).(cosx – cosxcosh + sinxsinh)/hcos(x+h)cos(x)
⇒ f'(x) = limh→0 (sec(x+h) + sec(x)).(cosx(1 – cosh) + sinxsinh)/hcos(x+h)cos(x)
Now, putting limh→0(1-cosh)/h = 0 and limh→0(sinh)/h = 1, we get,
⇒ f'(x) = limh→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.
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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