Find the equation of the tangent to the curve.

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

I just need help with differentiating!

Question

Find the equation of the tangent to the curve.

y= x sqrt. (3x+1)
when x = 5

I just need help with differentiating!

anonymous · Answer

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

anonymous · Answer

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?

callisto · Answer

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

callisto · Answer

v' = 1

.sam. · Answer

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

anonymous · Answer

Why wouldn't it be 

u= sqrt {3x+1} ----> u' = (1/2)(1/ sqrt 3x+1)
v= x -----> v' = 1

?

callisto · Answer

attachment

anonymous · Answer

Ok, but why times by 3?

anonymous · Answer

@order , chain rule:
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