Calculus question-finding second derivative involving trigonometry functions (should be easy)?

1. h(t) = 4 sin t - 5 cos t

The derivative of sin t is cos t, and the derivative of cos t is -sin t:

h ' (t) = 4 cos t - ( -5 sin t) = 4 cos t + 5 sin t

Using the same two derivative rules as before and taking the derivative again:

h " (t) = -4 sin t + 5 cos t

2. g(t) = 3 tan t

The derivative formula for y = tan x is dy/dx = sec^2 (x). I will derive it for you now using the quotient rule just so you can see how to derive the other three trigonometric formulas if you don't happen to have them memorized (you should have them memorized!):

y = tan x = sin x / cos x

dy/dx = [ (cos x)(cos x) - (sin x)(- sinx) ] / (cos x) ^2

dy/dx = [cos ^2 (x) + sin^2 (x) ] / cos ^2 (x)

dy/dx = 1 / cos ^2 (x)

dy/dx = sec^2 (x)

Getting back to the problem then:

g ' (t) = 3 sec ^2 (t)

For the second derivative you need to use chain rule, and you need to know that the derivative of sec t is (sec t)(tan t):

g " (t) = 3 (2 sec t) (sec t * tan t)

g " (t) = 6 sec ^2 (t) tan t

(When you try to simplify it, the trig gives you sine / cosine ^3, which can't really be made any simpler than the way it is now, so I leave the answer in its form).

It's just a matter of memorizing the appropriate formulas, being able to derive them from scratch if you don't happen to remember them, and being able to use these formulas in conjunction with the power rule, product rule, quotient rule, and chain rule- hopefully you know these already.

Finding Second Derivative

d/dt(u) means the first derivative of u with respect to t.

d^2/dt^2(u) means the second derivative of u with respect to t.

Is that what you mean by the temrinology I used that you don't understand?

OK, the first one is pretty easy. The first derivative of sin t = cos t . The first derivative of cos t is -sin t . So d/dt(4 sin t - 5 cos t) = d/dt(4 sin t) + d/dt(-5 cos t) = (4 cos t + (-5)*(-sin t) )dt = 4 cos t dt + 5 sin t dt . Make sense?

Then the second derivative d^2/dt^2(4 sin t - 5 cos t) = d/dt(4 cos t) + d/dt(5 sin t) = -4 sin t dt + 5 cos t dt . Understand? If not, add details to your question and I'll try to be more explicit.

For the second one, g(t) = 3 tan t = 3 (sin t / cos t) . Using the formula d(u/v) = ((v du) - (u dv))/v^2 ,

d/dt3(sin t / cos t) =

= 3(cos t * d/dt(sin t) - sin t * d/dt(cos t)) /cos^2 t

= 3(cos t * cos t) - sin t (-sin t)) dt / cos^2 t

= 3(cos^2 t + sin^2 t) dt /cos^2 t

Recall from trigonometry that sin^2 t + cos^2 t =1, so

3(cos^2 t + sin^2 t) dt /cos^2 t

= 3/cos^2 t = 3sec^2 t

Next I am again using the chain rulebelow to get the second derivative

d^2/dt^2(3 tan t) = d/dt(3 sec^2 t) . where I set u(t) = sec t

and I'm going to determine d/dt(3u^2) = 2*3*u * du/dt =

6sec t * d/dt(sec t) :

OK? Now all I have to do is determine d/dt(sec t) or the first derivative of sec t and multiply it by 6 sec t (I'll do that last, after I figure out d/dt (sec t)

By definition, sec t = 1/cos t = cos^(-1) t ,i.e. cos t raised to the -1 power. This is another chain rule application, where u = cos t and du = - sin t:

d/dt(sec t) = d/dt(1/cos t) = d/dt (cos^(-1) t)

= -1* cos^(-2) * (-sin t)

= (-1)*(-sin t) * cos^(-2) t = sin t / cos^2 t = (sin t / cos t) * sec t

= tan t sec t

Now, I multiply d/dt(sec t) by 2 * 3sec t * d/dt(sec t) as I said was going to do above (to complete application of the chain rule),

2 * 3sec t * d/dt(sec t) = 6 sec t * sec t tan t dt = 6 sec^2 t tan t dt

So d^2/dt^2(3 tan t) or the second derivative of 3 tan t =

6 sec^2 t tan t dt

Does that help at all?

By the way, I still say there's no dumb question, except possibly when you ask the same question right after I already answered it.

I know how you feel about calculus..but i actually understand this part of the Derivitive.

1.) h(t) = 4sin t - 5cos t

- you should know that the derivitive of:

sin(t) = cos(t)

cos(t) = -sin(t)

So from that..the answer of your problem is:

4cos t - 5(-sin t)......just change the sin and cos

but to the find 2nd derivitive..you must find the derivitive again of cos and -sin

= 4-sin t - 5(-cost)

2.) g(t) = 3 tan t

- same concept here..you just have to figure out what the derivitive of Tan equals.

so...you should know that Tan = Sin/Cos.

Since (Sin/Cos) is division..you need to use the quotient rule:

(sin)' (cos) + (sin) (cos)' all divided by (cosx)2

= (1/cos^2x) = sec x ^2

So the answer is...3 secx^2

h(t) = 4sint - 5cost

to get to the second derivative, you have to differentiate the above equation 2 times.

so, the first derivative is:

h'(t) = 4cost - (-5sint)

h'(t) = 4cost + 5sint

now, find the derivative of the above equation.

so, the second derivative is:

h''(t) = -4sint + 5cost

ok, now for the second problem, you have to remember that the derivative of tan t is 1/(cos² t)

therefore:

g(t) = 3tant

g'(t) = 3/(cos²t)

you could also write the above answer as: g'(t) = 3sec² t

since sect = 1/cost

1. h'(t)=4 cos t +5 sin t

h''(t)=-4sin t +5 cos t

Just use d/dx (sin x)=cos x & d/dx(cos x)=-sin x

2.g'(t)=3 (sec x)^2 Just find the derivative of tan t

Multiply the coeffecient by the exponent to get the new coefficient, then subtract one from the exponent to get the new exponent. Constants become 0 and cancel out. first derivative: f'(x)=18x²+9 second derivative: f''(x)=36x