Derivative of $\sec\left(\frac{3\pi}{2}-x\right)$ at $x=\pi/4$

Question

I was presented with the following excersice: Calculate the derivative of the following function and evaluate it at $x=\pi/4$

$\sec\left(\frac{3\pi}{2} - x\right)$

I used two different approaches and I was hoping for both to obtain the same results. Nevertheless this was not the case. Please find below my approaches

Approach 1: Using directly the quotient rule and substitute for x, meaning the following:

Step 1): Since we know that sec(x) = $\frac{1}{cos(x)}$, then we can apply the quotient rule directly and obtain the following: $\frac{\sin(x)}{\cos^2(x)}$

Step 2) Substitute for x --> $\left(\frac{3\pi}{2} - x\right) \rightarrow \left(\frac{3\pi}{2} - \frac{\pi}{4}\right)$ = $\frac{5\pi}{4}$

Step 3) Solve --> $\frac{\sin\left(\frac{5\pi}{4}\right)}{\cos^2\left(\frac{5\pi}{4}\right)}$ = $-\sqrt{2}$

Approach 2: Using the chain rule: When using the chain rule the result is the following:

$\sqrt{2}$

From what I have heard the right approach should be using the chain rule (approach 2)

Now since I do this more for the sake of learning instead of passing an exam or similar I would like to understand:

1-Why in this case the approach 1 is not suitable? (Otherwise I would have got the same results, right? Unless there is another kind of error, e.g.: alegebraic error or similar)

My assumption is that I am overseeing something when it comes to my conceptual understanding.

Answer 1

The approach No. 1 was false from the very beginnning since the composite function was ignored and thus both approaches are not equivalent.

Let $f(x)=\frac{3\pi}2-x.$ $$\begin{align}(\sec\circ f)'&=(\sec'\circ f)~f'\\&=-\sec'\circ f\\&\ne\sec'\circ f.\end{align}$$

Answer 2

In this particular simple case, it is useful to think that tho composite function has the effect of changing the direction of the $x$ axis (and shifting the origin, but that is not important for this problem). So if you look at the same point from two directions, a positive slope in one case will look like a negative slope in the opposite one.

Answer 3

$\begin{array}{rl}\text{Your first attempt:}\\\left.\dfrac{\mathrm d \sec((3\pi/2)-x)}{\mathrm d x}\right\vert_{x:=\pi/4}&\color{blue}{\neq}~\left.\left.\dfrac{\mathrm d \sec(y)}{\mathrm d y}\right\rvert_{y:=(3\pi/2)-x}\right\rvert_{x:=\pi/4}&\color{red}{\large\mathcal X}\\&=\left.\dfrac{\mathrm d \sec(y)}{\mathrm d y}\right\rvert_{y:=5\pi/4}\\[2ex]\text{Your second attempt:}\\\left.\dfrac{\mathrm d \sec((3\pi/2)-x)}{\mathrm d x}\right\vert_{x:=\pi/4}&=\left.\left(\left.\dfrac{\mathrm d \sec(y)}{\mathrm d y}\right\vert_{y=(3\pi/2)-x}\dfrac{\mathrm d~(3\pi/2)-x}{\mathrm d x}\right)\right\vert_{x:=\pi/4}&\color{green}{\large\checkmark}\\&=-\left.\dfrac{\mathrm d \sec(y)}{\mathrm d y}\right\rvert_{y:=5\pi/4}\end{array}$

1-Why in this case the approach 1 is not suitable? (Otherwise I would have got the same results, right? Unless there is another kind of error, e.g.: alegebraic error or similar)

You substituted into the derivand before you performed the differentiation. This would allow you to differentiate everything to 1:

$$\begin{align}\left.\dfrac{\mathrm d (x^3+x)^2}{\mathrm d x}\right\vert_{x:=2}&\neq\left.\dfrac{\mathrm d x^2}{\mathrm d x}\right\vert_{x:=2^3+2}\\&\neq\left.\dfrac{\mathrm d x}{\mathrm d x}\right\vert_{x:=(2^3+2)^2} \\&= 1\end{align}$$

So do not do that. Apply the Chain Rule:

$$\begin{align}\left.\dfrac{\mathrm d (x^3+x)^2}{\mathrm d x}\right\vert_{x:=2}&=\left.\dfrac{\mathrm d y^2}{\mathrm d y}\right\vert_{y=2^3+2}\left.\dfrac{\mathrm d x^3+x}{\mathrm d x}\right\vert_{x:=2}\\&=2(2^3+2)\cdot(3(2)+1)\\&=140\end{align}$$