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### Course: Calculus 1 > Unit 3

Lesson 2: More chain rule practice- Derivative of aˣ (for any positive base a)
- Derivative of logₐx (for any positive base a≠1)
- Derivatives of aˣ and logₐx
- Worked example: Derivative of 7^(x²-x) using the chain rule
- Worked example: Derivative of log₄(x²+x) using the chain rule
- Worked example: Derivative of sec(3π/2-x) using the chain rule
- Worked example: Derivative of ∜(x³+4x²+7) using the chain rule
- Chain rule capstone
- Proving the chain rule
- Derivative rules review

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# Worked example: Derivative of sec(3π/2-x) using the chain rule

We explore the fascinating process of differentiating the function sec(3π/2-x) in this worked example. Using the chain rule and trigonometric identities, we calculate the derivative and evaluate it at x=π/4. This problem illuminates the beauty of composite functions and their derivatives.

## Want to join the conversation?

- Why didn't Sal take the derivative of sec x as tan(x)sec(x)?(17 votes)
- He didn't take the derivative of sec x as tan(x)sec(x) because it is easier to input values into the sin and cos functions without a calculator. Calculating values of cosine are easier than calculating values of secant(30 votes)

- Why is not allowable to avoid the chain rule by letting y = sec (3*pi/2 - pi/4)

Then y = sec (5*pi/4)

Then y' = sin (5*pi/4) / (cos (5*pi/4)^2)

This evaluates to negative square root of 2...

Which is wrong! The answer is positive square root of 2.

But why does it come out wrong?

Why is it not valid to subtract the pi/4 at the beginning?(12 votes)- You are allowed to combine. However, in the video it is y=sec(3π/2 - x) and you are asked to evaluate dy/dx at x=π/4. What this means is you must differentiate first before substituting x=π/4 to evaluate.

You'll probably ask why. Because if you substitute x=π/4 in before differentiating (like you did) then the equation is not the same anymore and so will its derivative.(23 votes)

- At the very end, Sal evaluates:

-((-(√2)/2)/((-(√2)/2))^2) to √2. If I follow his method, I get the same result.

However, if I try to write it as -((-(√2)/2)/(*(-(√2)/2))*(-(√2)/2))*) and cancel out the top -(√2)/2) against one of the bottom -(√2)/2), i get 2/√2. I can't seem to figure out why...(10 votes)- If you multiply your result by √2/√2 (which is basically multiplying by 1), you'll get 2√2/(√2)^2 = √2.(16 votes)

- i got negative root 2. I just said deriv.(sec 3pi/2 - x) = sec(3pi/2 - x) * tan(3pi/2 - x) = (1) / (cos 5pi/4) * tan (5pi/4) if x=pi/4. then you get that tan 5pi/4 is 1. And, cos 5pi/4 is NEGATIVE root 2 / 2. 1 divided by that is - root 2.(6 votes)
- You forgot the (-1) from the derivative of the inside function.(14 votes)

- Can't the derivative of secx also be written secxtanx?(5 votes)
- Yes, the derivative of secx, which is sinx/(cos^2 x), can also be written secxtanx, because sinx/(cos^2 x) = (1/cosx) * (sinx/cosx) = secxtanx.(10 votes)

- What about using the trig identity sec((3pi/4)-x) = csc(x) ? You will get the same answer and you will get the answer faster.(3 votes)
- yes it is fine, but he's trying to use chain rule in here, beause it will be used through out the chapter.(4 votes)

- How about to differentiate y=(1-3x)^cosx by using implicit derivtive ?(3 votes)
- You would have to take the natural log of both sides and then differentiate implicitly from there

ln(y)=cos(x)*ln(1-3x)

1/y(dy/dx)= (-sin(x)*ln(1-3x)+(cos(x)*(-3/(1-3x))

dy/dx= y((-sin(x)*ln(1-3x)+(cos(x)*(-3/(1-3x))

You would then replace y with (1-3x)^(cos(x))`dy/dx= (1-3x)^(cos(x))*((-sin(x)*ln(1-3x)+(cos(x)*(-3/(1-3x))`

(4 votes)

- I just thought of something isn't sec(x) equal to 1/cos(x); then wouldn't that mean that the equation is a function of a function of a function, I mean 1/cos(x) is a composite function 1/x and cos(x) so I did the math and it is true(3 votes)
- Yes, you could consider sec(x) as a composite function. The same strategy can be used to find the derivative of sec(x) which sec(x)tan(x). We get the same result on differentiating 1/cos(x).(3 votes)

- I plotted the same function in desmos and it didn't seem to have a constant slope of sq-rt(2).(2 votes)
- √2 is the slope of the tangent line at 𝑥 = 𝜋∕4

For any other 𝑥, the slope is

−sin(3𝜋∕2 − 𝑥)∕cos²(3𝜋∕2 − 𝑥)

Follow this link to see how different values of x changes the tangent line:

www.desmos.com/calculator/m7jdexpoxh(4 votes)

- what about trying to differentiate f(x)=sinsqrtx ?

I am struggling because i am in scotland so we seem to write things differently from you guys, i have not yet heard of sec or csc or whatever; so far it's just been 1/cosx etc..

also, the book i am using has an annoying amount of wrong answers so its not advised to just work your way backwards from the answer since it could be wrong and you could be teaching yourself a wrong way of doing things (i've fallen n the trap before).

thanks(2 votes)- I understand that this was written 6 years ago, and if you know how to do this by now, just ignore this explanation.

If you mean sin(√x), then you would use the chain rule:

f(x) = sin(g[x])

g(x) = √x**or**x^(1/2)

f'(x) = cos(g[x])

g'(x) = 1/(2√(x))

According to the Chain rule, it would be f'(g[x]) * g'(x):

d(sin[sqrt{x}])/dx = cos(√x)/(2√x)(1 vote)

## Video transcript

- [Voiceover] So let's say
that we have y is equal to the secant of three pi over two minus x. And what we wanna do is,
we wanna figure out what dy/dx is, the derivative
of y with respect to x is, at x equal pi over four. Like always, pause this
video and see if you could figure it out. Well as you can see here, we
have a composite function. We're taking the secant, not just of x, but you could view this
as of another expression that x could define, or as
of a, of another function. So for example, if you call
this right over here, u of x, let's do that. So if we say u of x is equal
to three pi over two minus x, we could also figure out u prime of x is going to be equal to
derivative of three pi over two, that's just going to be a zero, derivative of minus x,
well that's just gonna be minus one, and you could just
view that as a power rule. It's one times negative one,
times x to the zero power, which is just one. So there you go. So we could view this as
the derivative of the secant with respect to u of x, that
when we take the derivative, the derivative of secant
with respect to u of x, times the derivative
of u with respect to x. And you might say, "Well, what about the
derivative of secant?" Well in other videos, we
actually prove it out, and you could actually re-derive it. Secant is just one over cosine of x, so it comes straight
out of the chain rule. So, in other videos, we prove that the, the derivative of the secant of x, of the secant of x, is equal
to, is equal to sin of x, over cosine of x, over
cosine of x squared. So if we're trying to
find the derivative of y with respect to x, well it's
going to be the derivative of secant with respect to u of x, times the derivative
of u with respect to x. So let's do that. The derivative of secant
with respect to u of x, well, instead of seeing an x everywhere, you're gonna see a u of x everywhere. So this is going to be, sin of u of x, sin of u of x, and I could, I don't have to write u of x, I could write three pi over two minus x, but I'll write u of x
right over here just to really visualize what we're doing. So sin of u of x, over, over, cosine squared of u of x, cosine squared, and we do those parentheses
in the blue color just to make sure that you identify it with the trig function. So cosine squared of u of x, u of x, so that's the derivative of
secant with respect to u of x, and then the chain rule
tells us it's gonna be that times u prime, u prime of x. So what is this going to be equal to? Well, I could just substitute back. This is going to be equal to,
I will write it like this. Sin of u of x, which is
three pi over two minus x, and I'll fill that in in a second, over cosine of u of x squared, times u prime of x, u of x is three pi over two minus x, three pi over two minus x. And then u prime of x,
we already figured out, is negative one, so I could write, times negative one, oh, yeah. Let me just leave it out there for now. I could have just put
a negative out front, but I really want you to be able to see what I'm doing here. And now we want to evaluate
at x equals pi over four. So that is equal to pi over four. Pi over four. So let's see, this is going to be, this is going to be equal to sin of, what's three pi over
two minus pi over four? I'll do that over here. So we have a common denominator, that is six pi over four, the same thing as three pi over two, minus pi over four, minus pi over four, is equal to five, five pi over four. So it's sin of five pi over four, five pi over four, over cosine squared of five pi over four, and then times negative one,
I can just put that out here. Now what is sin of five pi over four and cosine squared of five pi over four? Well, I don't have that memorized, but, let's actually draw a unit circle and we should be able to, we should be able to
figure out what that is. So, a unit circle, I try to hand draw it as best as I can. Please forgive me that
this circle does not look really like a circle, alright. Okay. So, let me just remember my angles. So, I, I in my brain, I
sometimes convert into degrees. Pi over four is 45 degrees. This is pi over two, this
is three pi over four, this is four pi over four, this is five pi over four, lands you right over there. So if you wanted to see where
you intersect the unit circle, this is at the point,
this is at the point. Your x-coordinate is negative
square root of two over two, negative square root of two over two, and your y-coordinate
is negative square root of two over two. If you are wondering how I got that, I encourage you to review the unit circle and some of the standard
angles around the unit circle. You'll see that in the
trigonometry section of Khan Academy. But this is enough for us, because the sin is the y-coordinate, it's
the y-coordinate here. So negative square root of two over two. So this is negative square
root of two over two. And then, the cosine is the x-coordinate, which is also negative
square root of two over two, it's going to be that squared. Negative square root of two
over two, we're squaring it. So if we square this, it's gonna become, this is going to become,
it's gonna become positive. And then square root
of two squared is two, and then two squared is four. So it's one half. So this is, the denominator's
equal to one half. See the numerator, this negative cancels out with that negative. And so we are left with, we deserve a little bit of a drum roll. We are left with square
root of two over two, that's the numerator, divided by one half, well
that's the same thing as multiplying by two. So we are left with
positive square root of two, is the slope of the tangent line, to the graph of y is equal to this when x is equal to pi over four. Pretty exciting.