Using the definition of the derivative at x=c, compute the following derivatives at a

• Feb 1st 2009, 01:57 PM
jkami
Using the definition of the derivative at x=c, compute the following derivatives at a
Using the definition of the derivative at x=c, compute the following derivatives at an arbitrary point?

(a) g(x) = 5x^2

(b) f(x) = 7x^4

(c) k(x) = sqrt x

Please show me the steps cuz I want to learn, not just copying the answer
• Feb 1st 2009, 02:06 PM
Plato
$\frac{{k(x + h) - k(x)}}
{h} = \frac{{\sqrt {x + h} - \sqrt x }}
{h} = \frac{{\sqrt {x + h} - \sqrt x }}
{h}\frac{{\sqrt {x + h} + \sqrt x }}
{{\sqrt {x + h} + \sqrt x }} = \frac{h}
{{h\left[ {\sqrt {x + h} + \sqrt x } \right]}}$

Now divide out the h's and find the $\lim _{h \to 0}$
• Feb 1st 2009, 02:32 PM
jkami
Quote:

Originally Posted by Plato
$\frac{{k(x + h) - k(x)}}
{h} = \frac{{\sqrt {x + h} - \sqrt x }}
{h} = \frac{{\sqrt {x + h} - \sqrt x }}
{h}\frac{{\sqrt {x + h} + \sqrt x }}
{{\sqrt {x + h} + \sqrt x }} = \frac{h}
{{h\left[ {\sqrt {x + h} + \sqrt x } \right]}}$

Now divide out the h's and find the $\lim _{h \to 0}$

I know this formula, but I don't understand what does x = c mean.

so, the first one I got lim g(x) = 10x, what do i have to do next
• Mar 1st 2009, 12:41 AM
CaptainBlack
Quote:

Originally Posted by jkami
I know this formula, but I don't understand what does x = c mean.

so, the first one I got lim g(x) = 10x, what do i have to do next

No you go g'(x)=the limit=10x, that is the deriviative at the point x is 10x. x is a dummy variable (the logicians have another name for this but I will not bother you with that) it can be replaced with any other symbol denoting a variable, so you also have g'(c)=10c etc.

CB