Question

take derivative of y= square root of (2x+1) by using limit definition?

Answer 1

y'(x) = 1/sqrt(2 x+1)

Answer 2

i know how to do derivative, but ca you show me the step for using limit definition?

Answer 3

\[y \prime(x)=\lim_{h \rightarrow 0} (y(x+h)-y(x))/h)\]

Answer 4

\[=\lim_{h \rightarrow 0} ((\sqrt{2(x+h)+1}-\sqrt{2x+1})/h)\]
multiply by the conjugate
\[=\lim_{h \rightarrow 0} ((\sqrt{2(x+h)+1}-\sqrt{2x+1})(\sqrt{2(x+h)+1}+\sqrt{2x+1})/h(\sqrt{2(x+h)+1}+\sqrt{2x+1})\]
\[=\lim_{h \rightarrow 0} ((2(x+h)+1-2x+1)/h(\sqrt{2x+2h+1}+\sqrt{2x+1}))\]
\[=\lim_{h \rightarrow 0} (2h/h(\sqrt{2x+2h+1}+\sqrt{2x+1}))\]
\[=\lim_{h \rightarrow 0} (2/(\sqrt{2x+2h+1}+\sqrt{2x+1}))\]
\[=2/(\sqrt{2x+2(0)+1}+\sqrt{2x+1})\]
\[=2/(2\sqrt{2x+1})\]
\[=1/\sqrt{2x+1}\]

Answer 5

oops on the 3rd line i forgot brackets around the -2x+1 should be -(2x+1)

Answer 6

thank you so much!