Here's the question you clicked on:
coco1234
Suppose f '(2) = 3 and g '(2) = 8 Find h'(2) where h(x) = 5f(x) + 3g(x) +1 h'(2) =
h'(x) = 5f'(x) +3g'(x) +1 => h'(2) = 5(3) +3(8) +1 =15+24+1 = 40
the answer to the problem said that it was 39:PP its like an online hw assignment
i dont know how its 39 though
ohhhhhhhhhhh i'am sorry i made a mistake!!!
h'(x) = 5f'(x) +3g'(x) +1 => h'(2) = 5(3) +3(8) =15+24 = 39 (if we differentiate 1 we get 0)
ohhh i see haha thank you:)
h[x] = 5 f[x] + 3 g[x] + 1 h'[2]=5 f ' [2]+3 g'[2] h'[2]=5*3 + 3*8 h'[2]=15 + 24 h'[2]=39
myinmi, hello, i think i see why the discrepancy in my problem
in the problem you have to take the square root of tan^2 theta, but we know that the square root of x^2 is |x|, and for the theta in the problem we have a negative answer