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
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.)
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