

Trig in Calculus
y = sin x y’ = cos x y = cos x y’ = sin x y = tan x y’ = sec^{2} x These are rules that you will have to follow. They are also very basic, and can be complexified in interesting ways. (Did I just create a word?) In the spirit of having you actually remember some of the above, I will allow you to see how the derivative of sin x is cos x. Carefully draw both a sin and cos curve on one graph, from 0 to 2∏. (That symbol is pi) You will notice at x = 0 that the angle that the sin curve goes through the origin is a 45 degree slop upwards, which means that it goes up 1 y for every 1 x. The slope is then 1, and the derivative at that point should be 1. Cos of 0 is 1. At x = ∏/2, sin x reaches a maximum, at which the slope is 0. Cos ∏/2 is 0. At ∏, the sin curve has a 1 slope, and cos is 1. This pattern is correct not only at these points but at every point. A similar analysis will show that the derivative of cos x is sin x. With complex rulesy = 5sin x y’ = 5cos x This is because the 5 is a coefficient, and any time you are taking a derivative of a function with a number in front, you can take that number away for the purposes of taking the derivative, and put it back in after solving. y = sin 5x y’ = 5cos 5x This is simply the chain rule, used in the last topic. The 5x is u and the derivative of sin u is cos u. The derivative of 5x is 5. Multiply the derivative of outside by derivative of inside, and you have the derivative of equation. Then you replace the u. y = 3sin 5x y’ = 15cos 5x y = 3cos 5x y’ = 15sin 5x y = sin^{2} x = (sin x)^{2} = u^{2 }(and u = sin x) y’ = 2u(cos x) = 2sin x cos x This one I solved by a chain rule similar to what you saw in the chain rule section. Notice that I put the square around the entire sin x, this is explained in the Intro. Now I made sin x the inner function, and u^{2} the outer function. The rest is simple. Take the derivatives of both and multiply. Then replace the u to its original status (sin x). y = 3sin^{2} (5x + 3) This one is not as tough as it looks if you understood everything before. On the other hand, if you understood everything before it probably doesn’t look too tough. This requires a chain rule inside a chain rule. The equation rearranges as 3[sin (5x + 3)]^{2} The outside equation to solve is u^{2} and u = sin (5x +3). So If I take the derivative of the outside, getting 2u, and multiply it by the derivative of the inside function, then I have my answer. But what is the derivative of the inside function? The derivative of sin (5x +3) requires a chain rule. Once you figure out the derivative, take it back and multiply it by the other one. The derivative of this is 5 * cos (5x + 3). Multiply that by 2u, and multiply by the 3 that we've been ignoring, and you have 30u cos (5x + 3) which is 30[sin (5x + 3)]cos (5x + 3). I hope I didn’t screw that up. No, I think it’s right. y = 2tan^{4} x y’ = 8tan^{3} x sec^{2} x This is nothing you can’t handle. The u turned back into a tan x at the end, and 8u^{3}sec^{2}x became 8tan^{3} x sec^{2} x. y = sin (ln x) y’ = cos (ln x) * 1/x y = ln (sin x) y' = (1/sin x) * (cos x) = cos x/sin x = cot x y = ln (cos x) y' = (1/cos x) * ( sin x) =  sinx/cos x =  tan x y = cos (ln 2x) y’ = sin (ln 2x) * 1/x I used two chains in that one. The 2x must become a u in order to get the derivative of a ln 2x. The following example uses a derivative that you are required to know, and is shown along with the rest in my reference tables. y = tan x/sin x = sec x y' = sec x tan x y = tan (e^{x}x^{2}) y’ = sec^{2} (e^{x}x^{2}) * (e^{x}x^{2} + e^{x}2x) That one has a product rule inside the chain rule. Make sure your sec^{2} is left with a (e^{x}x^{2}) inside, because during the solving process it turns into a u. OK, I think I’ve done enough trigonometries for one day. I’ll move on to implicit differentiation.
Prev Lesson  Problem Sets  Next Lesson  Index of Topics  Reference

