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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 26: Trigonometric functions differentiation

# Derivative of tan(x) (old)

An older video where Sal finds the derivative of tan(x) using the quotient rule. Created by Sal Khan.

## Want to join the conversation?

• Question on that last step - couldn't you also cancel out cos^2 x in the numerator and denominator, making the solution sin^2 x?
• You can cancel it out with the cos^2x, but you won't end up with just sin^2x, because what you actually have is: cos^2x/cos^2x + sin^2x/cos^2x. Which ends up being 1 + sin^2x/cos^2x or 1+tan^2x = sec^2x.
• I tried this using a product rule, but I couldn't get same result. Is this only achieveable by quotient rule?
• Well as Sal said, the quotient rule is a derivation of the product rule so yes you most definitely can do it through the product rule. It's a bit trickier though and you have to go through the working in the right way. Quotient rule is much easier =)
• would the derivative of tan 5x then be 5sec^2 5x?
• Yessir.
• At , how does cos^2(x)+sin^2(x) = 1? And at , how does the negative sign become a positive sign with -sin(x)?
(1 vote)
• what would be the differentiation of 4 tan^2 5X +3

need hepl
any one there to answere this question
• Just for practice, I tried to derive d/dx(tanx) using the product rule. It took me a while, because I kept getting to (1+sin^2(x))/cos^2(x), which evaluates to sec^2(x) + tan^2(x). Almost there, but not quite. After a lot of fiddling, I got the correct result by adding cos^2(x) to the numerator and denominator. I'm just curious, can anyone tell me if this is the typical move? Is there another way to derive this using the product rule?
• I think you made a mistake somewhere. Here is what I got.
tan x = sin x sec x
d/dx tan x = d/dx (sin x sec x)
d/dx tan x = sin x ( sec² x * sin x) + cos x (sec x)
d/dx tan x = (sin² x )(sec² x) + 1
d/dx tan x = (sin² x / cos² x) + 1/1
The LCD is cos²x
d/dx tan x = sin²x/cos² x + cos²x/ cos²x
d/dx tan x = (sin² x + cos² x) / cos² x
d/dx tan x = 1 / cos² x
d/dx tan x = sec² x

Note: to get d/dx (sec x) I did not look up a value, I did this:
d/dx (sec x) = d/dx (cos x)⁻¹
= (−1) (cos x)⁻² d/dx (cos x)
= (−1) (cos x)⁻² (- sin x)
= sin x / cos² x
= (sec² x)( sin x)
• So, similarily, the derivative of tanx is equal to 1+(tanx)^2

that's a neat result
• indeed it is. if you take the expression

sin(x)^2 + cos(x)^2 = 1

and divide everything by

cos(x)^2

you get

(sin(x)^2)/cos(x)^2 + 1 = (1/cos(x)^2)

which using the other trigonometric identities can be simplified to

tan(x)^2 + 1 = sec(x)^2
• So this may be a rather stupid Question, but why is the derivative of tan(x) = to sinx/cosx? it may have been asked below but im still confused. Doesn't the tan(x) = sec2(x)
(1 vote)
• We have quite a few "identities" for trig functions, and one of the most important ones is:

tanx = sinx/cosx

So this is not the derivative of tanx. Instead, it's just a different way of writing tanx. If you don't recognize this identity, it might be a good idea to spend some time viewing the videos dealing with trig identities.

The derivative of tanx is sec²x.
• Could someone do it using the chain rule, please?
(1 vote)
• d/dx[tan(x)]
= d/dx[sin(x) / cos(x)]
= d/dx[sin(x) * 1/cos(x)]
= d/dx[sin(x) * cos(x)^-1]
= d/dx[sin(x)](cos(x))^-1 + (sin(x))(d/dx[cos(x)^-1]) # product rule
= cos(x)cos(x)^-1 + sin(x)(-cos(x)^-2)(-sin(x)) # chain rule
= cos(x)/cos(x) + sin(x)^2/cos(x)^2
= cos(x)^2/cos(x)^2 + sin(x)^2/cos(x)^2
= (cos(x)^2 + sin(x)^2) / cos(x)^2
= 1 / cos(x)^2
= sec(x)^2
• That is an awful lot of explanation for such a simple answer, but I can see how knowing exactly each step, will expand my knowledge of the concepts and principles needed to further my calculus ability.