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

Lesson 26: Trigonometric functions differentiation- Derivatives of tan(x) and cot(x)
- Derivatives of sec(x) and csc(x)
- Derivatives of tan(x), cot(x), sec(x), and csc(x)
- Worked example: Derivative of sec(3π/2-x) using the chain rule
- Differentiate trigonometric functions
- Derivative of tan(x) (old)
- Differentiating trigonometric functions review

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# 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?(13 votes)
- 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.(25 votes)

- I tried this using a product rule, but I couldn't get same result. Is this only achieveable by quotient rule?(4 votes)
- 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 =)(7 votes)

- would the derivative of tan 5x then be 5sec^2 5x?(4 votes)
- At2:53, how does cos^2(x)+sin^2(x) = 1? And at1:56, 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(2 votes) - 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?(2 votes)
- 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)(3 votes)

- So, similarily, the derivative of tanx is equal to 1+(tanx)^2

that's a neat result(2 votes)- 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(3 votes)

- 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.(4 votes)

- 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(4 votes)

- 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.(2 votes)
- Yes, but the answer is simple because we already know it from the work done by those mathematicians of years gone by.

And you are very very correct when you say "I can see how knowing exactly each step, will expand my knowledge of the concepts and principles needed to further my calculus ability."

As you get further into calculus, it is less and less a case of "this type of problem requires this type of solution" and more like, "given the properties and concepts you now know, how can you use your intuition to create a solution." You will see, that in one point of view, the more rigorous the math you learn, the more creative and "arty" you need to be with it, but without breaking the rigor by one iota! You will see some amazing leaps of creativity that have given rise to many of the solutions to the most formidable challenges in math.

Remember, everything, and I mean everything that you are learning in math is a structure to prepare and solidify your intuition to apply to the greater concepts to come, and these greater concepts are not the end, they themselves are structures for even greater things.

How far will you go?(2 votes)

## Video transcript

In the last video, we
saw that the quotient rule, which, once again, I have
mixed feelings about because it really comes straight
out of the product rule. If we have something in the
form f of x over g of x, then the derivative
of it could be this business right over here. So I thought I would at
least do one example where we can apply that. And can could do it to find the
derivative of something useful. So what's the
derivative with respect to x-- let me write
this a little bit neater-- the derivative with
respect to x of tangent of x? And you might say,
hey Sal, wait, I thought this was
about the quotient rule. But you just have
to remember, what is the definition
of the tangent of x? Or what is one way to
view the tangent of x? The tangent of x
is the same thing as sine of x-- let me
now color code it-- is the same thing as sine
of x over cosine of x. And now it looks clear that
our expression is the ratio or it's one function
over another function. So now we can just
apply the quotient rule. So all of this
business is going to be equal to the derivative of
sine of x times cosine of x. So what's the
derivative of sine of x? Well, that's just cosine of x. So it's cosine of x is
derivative of sine of x times whatever function we
had in the denominator. So times cosine of
x minus whatever function we had
in the numerator, sine of x, times the
derivative of whatever we have in the denominator. Well what's the
derivative of cosine of x? Well the derivative of cosine
of x is negative sine of x. So we'll put the sine of x here. And it's a negative
so I could just make this right over
here a positive. And then all of
that over, whatever was in the denominator
squared, all of that over cosine
of x squared. Now what does this simplify to? Well in the numerator
right over here, we have cosine of x
times cosine of x. So all of this simplifies
to cosine squared of x. And sine of x times sine of x,
that's just sine squared of x. And what's cosine squared
of x plus sine squared of x? This is one of the most basic
trigonometric identities. It comes straight out of
the unit circle definition of trig functions. Let me write it over here. Cosine squared of
x plus sine squared of x is equal to 1, which
simplifies things quite nicely. So cosine squared of
x plus sine squared of x, all of this entire
numerator is equal to 1. So this nicely simplifies to
1 over cosine of x squared, which we could also write like
this, cosine squared of x. These are two ways of
writing cosine of x squared, which is the same exact thing
as 1 over cosine of x squared, which is the same
thing as secant. 1 over cosine of x
is just secant of x. Secant of x squared, or we
could write it like this. Secant squared of x. And so that's where
it comes from. If you know that the
derivative of sine of x is cosine of x and the
derivative of cosine of x is negative sine of x,
we can use the quotient rule, which, once again, comes
straight out of the product rule to find the
derivative of tangent x is secant squared of x.