Question

question

Answer 1

If

\[f(x) =uv \]

evaluate it using \[\lim_{x \rightarrow 0} \frac{ f(x+\Delta~x )-f(x) }{ \Delta~x }\]

Answer 2

here is what I have so far

\[\frac{ u(x+\Delta~x)v(x+\Delta~x)-u(x)v(x) }{ \Delta~x}\]

Answer 3

\[f'(x)=\lim_{\Delta x\ \rightarrow 0}\frac{ f \left( x+\Delta x \right)-f \left( x \right) }{ \Delta x }\]
\[=\lim_{\Delta x \rightarrow 0}\frac{ \left( u+\Delta u \right)\left( v+\Delta v \right)-uv }{ \Delta x }\]
\[=\lim_{\Delta x \rightarrow 0}\frac{ \left( u+\Delta u \right)\left( v+\Delta v \right)-uv }{ \Delta x }
=\lim_{\Delta x \rightarrow 0} \frac{ \left( u+\Delta u \right)\left( v+\Delta v \right)-\left( u+\Delta u \right)v+\left( u+\Delta u \right)v-uv }{ \Delta x }\]
\[=\lim_{ \Delta x \rightarrow 0}\frac{ \left( u+\Delta u \right)\left( v+\Delta v-v \right)+\left( u+\Delta u-u \right)v }{ \Delta x }\]
\[=\lim_{\Delta x \rightarrow 0}\frac{ \left( u+\Delta u \right)\Delta v }{ \Delta x }+\lim_{\Delta x \rightarrow 0} \frac{( \Delta u) v } { \Delta x }\]
\[=\lim_{\Delta x \rightarrow 0}\left( u+\Delta u \right)\frac{ \Delta v }{ \Delta x }+\lim_{\Delta x \rightarrow 0}v \frac{ \Delta u }{ \Delta x }\]
when \[\Delta x \rightarrow0,\Delta u \rightarrow0,\Delta v \rightarrow 0,\frac{ \Delta u }{ \Delta x }\rightarrow \frac{ du }{ dx } \rightarrow~u',\frac{ \Delta v }{ \Delta x }\rightarrow \frac{ dv }{dx }\rightarrow v'\]
?