ap calc deriv review

[xtrmk]

If y = asin(ct) + bcos(ct), where a, b, c are constants, then y'' = ?

How do I start this problem?

If y = sqrt(x^2+16) what is y''?

I got y ' to equal (1/2)(x^2+16)^(-1/2) (2x)
after get it down to x/sqrt(x^2+16) , I tried using the quotient and product rule but cannot get a definite answer.

If H(x) = sqrt[(F(x)], then H'(3) = ?
Can someone help me to set up H because i have the given f, f', g , and g' on a chart.

thank you for any help.

 

[skeeter]

all of these are just chain rule problems ...

y = a*sin(ct) + b*cos(ct)

y' = ac*cos(ct) - bc*sin(ct)

y" = -ac²*sin(ct) - bc²*cos(ct)


y = (x² + 16)^1/2

y' = (1/2)(x² + 16)^-1/2(2x)

y' = x(x² + 16)^-1/2

y" = x[(-1/2)(x² + 16)^-3/2(2x)] + (x² + 16)^-1/2

y" = -x²(x² + 16)^-3/2 + (x² + 16)^-1/2

y" = (x² + 16)^-3/2[-x² + (x² + 16)]

y" = 16(x² + 16)^-3/2 = 16/(x² + 16)^3/2


h(x) = [f(x)]^1/2

h'(x) = (1/2)[f(x)]^-1/2f'(x) = f'(x)/(2[f(x)]^1/2)