For ex, I’m trying to intuitively understand why the derivative of secx = secxtanx as opposed to just memorizing it.

When it comes to the derivatives of trig functions, are we still looking at the slope of the tangent line of f(x)? Not sure if this question makes sense. Thanks!

In: 1

Yes, the derivative is still the slope of the function. [Here’s](https://www.mathsisfun.com/calculus/derivatives-trig-proof.html) a link to showing proofs of the derivatives of sine and cosine, and once you know those you can apply other derivative rules to find the other trig functions.

>When it comes to the derivatives of trig functions, are we still looking at the slope of the tangent line of f(x)?

Yes, you are. The interpretation of the derivative is the same regardless of what function you’re differentiating, although you’ll need to expand it a little bit once you’re dealing with functions of multiple variables (typically in a calc 3 class).

I’ve put up [a little Desmos graph to demonstrate](https://www.desmos.com/calculator/3ttc6n3bhk). On that graph, you can see, the tangent and secant together with a unit circle. You can see, that “tangent” is really a tangent to circle, and “secant” is where the tangent inter**sects** the X axis.

The question of derivative basically boils down to the question: if we make a little nudge `dx` to the `x`, how much does `sec x` grows? In other words, how much is `d(sec x)`? You can control `dx` on the graph with the first slider (labeled `ϕ_d`). You can also use the “play” button on the left of it to play an animation.

The change to `sec x` is comprised of 3 factors:

* Rotation of tangent line

* “Extension” of tangent line, until it intersects the secant.

* Movement of point P around the circle

Let’s consider them one by one:

**Step 0.** Movement of point P around the circle can be ignored. For small `dx`, its contribution to the result is negligible. We can just pretend, that tangent line just rotates in place.

**Step 1.** Let’s consider, that tangent line rotates, but doesn’t extend. Where the secant point S0 will be? Well, it rotates `dx` radians around P, on a circle of radius `tan x`. So, it moves away `tan x*sin dx` distance. Because `dx` is small, `sin dx ~= dx`, so we get `tan x * dx`. Notice, that movement is perpendicular to original tangent line. That means it parallel to radius OP (because OP is also parallel to tangent).

*Step 1 conclusion.* S0 moves distance `tan x * dx` parallel to OP.

**Step 2**. Now let’s consider, that tangent moves with S0 and extends, but doesn’t rotate. That means, we “move” tangent line `tan x * dx` away from the circle. Radius OP = 1, but now it grows to `1 + tan x * dx`, so it scales by a factor of `1 + tan x * dx`. That means, that secant OS should also grow by a factor `1 + tan x * dx` (from triangle POS, it’s angles don’t change, so it stays similar to itself). So `sec x + d(sec x) = sec x * (1 + tan x * dx) = sec x + sec x * tan x * dx`, which means `d(sec x) = sec x * tan x * dx`.

So, derivative of `sec x` is `d(sec x)/dx = sec x * tan x`. The `tan x` is from rotating the tangent line without extending, `sec x` is from moving and extending tangent line without rotating.

Not a five year old subject matter, but another way to investigate integrals and derivatives of trig functions is too express them as complex exponentials. This has many advantages since derivatives and integrals of exponential functions is particularly easy…

The wiki article on Euler’s Formula shows several useful formulas for this…

sec(x) = 1/cos(x), so you can use the chain rule to find its derivative.

Let u = cos(x), so sec(x) = 1/u = u^(-1).

Then the derivative with respect to x is -u^(-2) × d/dx u

Remembering that u = cos(x):

= -1/cos^(2)(x) × -sin(x)

= sin(x)/cos(x) × 1/cos(x)

= tan(x).sec(x)

You only need to memorise the derivatives of sin(x) and cos(x), then you can use them to determine the derivatives of the other trig functions using the chain and quotient rules.

(And yes, these are still formulae for the gradient of the tangent at point x.)