Question

Trouble grasping an implicit differentiation problem.

This problem (2E-7: http://goo.gl/ekme9) would have it that:

\[V=4h^2 + h/2\]
=>
\[V'=8hh'+2h'\]

if anyone could explain how we get the 2h' part I would be deeply obliged.
V

Answer 1

its like differentiation of x wrt y,
its made simple,
its like \[4(y^2 +\frac{ y }{ 2 })\]
\[4y^2 + 2y\]
now the differentiation of y^2 wrt x is 2y*dy/dx and of y = dy/dx
\[2(4y)^1 \times y \prime + 2y \prime\]
you understand?

Answer 2

I'm sorry for phrasing my question unclearly. I get that h is just a different variable here. And I understand how [\4h^2 => 8hh'\], what is still confusing me is the algebra behind [\h/2 => 2h'\]

How do we get the result 2h'.
Sorry for asking something this elementary.

Answer 3

not a different "variable", but rather, a function in its own right

Answer 4

\[V=4(h(t))^2 + \frac12h(t)\]
\[V'=[4(h(t))^2]' + [\frac12h(t)]'\]
now, dont forget to apply the chain rule
\[V'=4*2(h(t))*h'(t) + \frac12*h'(t)\]
if we do a lazy write up and know that h is a function of t
\[V'=8hh' + \frac12h'\]

Answer 5

the only way to get a "2" h' is if you posted the question incorrectly ...

Answer 6

Thanks for the answer amistre64, this means that there was actually a mistake in the solutions. Wanted to confirm this. You helped.

Answer 7

youre welcome :)