Question

find an equation for the tangent to the curve at the given point. y= (2) sqrt(x) (1,2)

Answer 1

dy/dx=gradient of tangent=1(x)^-1/2
at x=1 m= 1
equation is
y-2=1(x-1)
y=x+1

Answer 2

i have to do it with like the limit as h goes to 0.

Answer 3

yes that mean of diferentiate equation

Answer 4

lim as h goes to 0 [f(x+h) - f(x)] / h

Answer 5

like in that process?

Answer 6

its the same thing his way is just easier

Answer 7

\[\frac{dy}{dx} = \lim_{h \rightarrow 0} \frac{ 2\sqrt{x+h} - 2\sqrt{x} } {h} \]

Answer 8

factor out the 2, and multiply top and bottom by \[\sqrt{x+h} + \sqrt{x} \]

Answer 9

\[\frac{dy}{dx} = 2\lim_{h \rightarrow 0} \frac{h} {h(\sqrt{x+h}+\sqrt{x})} \]

Answer 10

\[\frac{dy}{dx} = 2 \lim_{h \rightarrow 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} \]

Answer 11

\[\frac{dy}{dx} = 2 \frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{x} } \]

Answer 12

etc, rest is same

Answer 13

ohh i see! thank you!