What's the first derivative of f(x) = tan(sqrt(3x^2+2x+2))? I'm getting sec^2(sqrt(3x^2+2x+2)) * 1/(2*sqrt(3x^2+2x+2)) * (6x+2) ... but that seems to be incorrect.

Question

lgbasallote · Answer

$$\huge f(x) = \tan(\sqrt{3x^2+2x+2})$$
is that the question?

anonymous · Answer

Indeed.

lgbasallote · Answer

well your answer would be right...next step would be to express 6x + 2 as 2(3x+1)

so you see something will cancel

anonymous · Answer

Thing is, I'm actually only being asked to find f'(1), but when I sub 1 into the equation I got there, I don't seem to get the answer I'm apparently supposed to be getting (which is 1.95)... instead, I get 1.52... I'm not sure what I'm doing wrong :\

anonymous · Answer

It might be the way you are entering it into your calculator.  I got the same derivative that you did in the original question, but when I tried to punch it into my calculator I got an error message, so I tried it again except I evaluated the contents inside the radical (with x=1 the inside is 7 i think), and I also evaluated 6x+2 as just 8.  When I put it into my calculator that way I got the same answer that you provided as the correct answer.  1.954210782

lgbasallote · Answer

$$\LARGE  f'(x) = \frac{(3x+1)(\sec^2 (\sqrt{3x^2+2x+2})}{\sqrt{3x^2 + 2x +2}}$$
so therefore
$$\Large f'(1) = \frac{(3(1) + 1) (\sec^2\left (\sqrt{3(1)^2 + 2(1) + 2}\right)}{\sqrt{3(1)^2+2(1) + 2}}$$
is that what you did?

anonymous · Answer

Are you asking me, or the who needed help?