Question

Find the equation of the tangent to the curve.

y= x sqrt. (3x+1)
when x = 5

I just need help with differentiating!

Answer 1

\[y= x \sqrt {(3x+1)}\]

Answer 2

Can you do it in the form
\[u = \sqrt {3x+1} ---> u' =\]\[v = x -----> v' = \]
It makes it easier for me to understand... What's the diferentiation of u?

Answer 3

u' =(1 /2) [sqrt (3x+1)] ^(-1/2) (3) = 3/2[sqrt (3x+1)] ^(1/2)

Answer 4

v' = 1

Answer 5

y=(copy left)(diff. right)+(copy right)(diff left)
\[y=x[ (1/2)(3x+1)^{-1/2} (3) ] + (\sqrt{3x+1})\]

Answer 6

Why wouldn't it be

u= sqrt {3x+1} ----> u' = (1/2)(1/ sqrt 3x+1)
v= x -----> v' = 1

?

Answer 7

Ok, but why times by 3?

Answer 8

@order , chain rule: