Now find the second derivative when x = 2 and y =1

I keep getting 3/4 as my answer. I solved it using implicit differentiation and also using partials differentiation to double check. The correct answer according to blackboard is -14/27

and can someone show me what im doing wrong with my latex. Guess im a bit rusty. Thanks

Aug 3rd 2011, 09:26 PM

pickslides

re: Second Derivative Value via Implicit Differentiation

Quote:

Originally Posted by 11rdc11

and can someone show me what im doing wrong with my latex. Guess im a bit rusty. Thanks

Use "tex" tags, "math" tags are no good here at the moment

Now find the second derivative when x = 2 and y =1

I keep getting 3/4 as my answer. I solved it using implicit differentiation and also using partials differentiation to double check. The correct answer according to blackboard is -14/27

and can someone show me what im doing wrong with my latex. Guess im a bit rusty. Thanks

Your function is defined by $\displaystyle \frac{y^{3}}{3} +2y -x^2 + \frac{5}{3} = 0 $ . Deriving: $\displaystyle y^2y'+2y'-2x=0$ . Substituting $\displaystyle x=2,y=1$ we get $\displaystyle 3y'-4=0$ or equivalently $\displaystyle y'=4/3$ . Derive again and substitute $\displaystyle x=2,y=1,y'=4/3$ .

Aug 3rd 2011, 09:48 PM

11rdc11

re: Second Derivative Value via Implicit Differentiation

Thanks guys but even when i take the second derivative i get the same

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

But this is wrapped inside the legs-uncrossed version of...