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

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.C (LO)
,
FUN‑3.C.1 (EK)
Sal differentiates sec(3π/2-x) and evaluates the derivative at x=π/4.

## Want to join the conversation?

• Why didn't Sal take the derivative of sec x as tan(x)sec(x)?
• 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
• 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?
• 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.
• 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...
• If you multiply your result by √2/√2 (which is basically multiplying by 1), you'll get 2√2/(√2)^2 = √2.
• 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.
• You forgot the (-1) from the derivative of the inside function.
• Can't the derivative of secx also be written secxtanx?
• 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.
• I plotted the same function in desmos and it didn't seem to have a constant slope of sq-rt(2).
• √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
• 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.
• yes it is fine, but he's trying to use chain rule in here, beause it will be used through out the chapter.
• How about to differentiate y=(1-3x)^cosx by using implicit derivtive ?
• 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))``