Question

derivative find dy/dx
y=(x+1)(x+2)/(x-1)(x-2)

Answer 1

use

(f g) ' = f ' g + g ' f

Answer 2

i cant do it

Answer 3

tell me the derivative of each:

1) x+1
2) x+2
3) x-1
4) x-2

Answer 4

:(

Answer 5

you can do like this also.
\[\ln y=\ln \frac{ \left( x+1 \right)\left( x+2 \right) }{ \left( x-1 \right)\left( x-2 \right)}\]
\[\ln y=\ln \left\{ \left( x+1 \right)\left( x+2 \right) \right\}-\ln \left\{ \left( x-1 \right)\left( x-2 \right) \right\}\]
\[\ln y=\ln \left( x+1 \right)+\ln \left( x+2 \right)-\ln \left( x-1 \right)-\ln \left( x-2 \right)\]
diff. w.r.t. x ,
\[\frac{ 1 }{ y }\frac{ dy }{dx }=\frac{ 1 }{x+1 }+\frac{ 1 }{ x+2 }-\frac{ 1 }{x-1 }-\frac{ 1 }{ x-2 }\]
\[\frac{ dy }{dx }=y \left\{ \frac{ 1 }{ x+1 }+\frac{ 1 }{ x+2 }-\frac{ 1 }{ x-1 }-\frac{ 1 }{x-2 } \right\}\]
\[\frac{ dy }{dx }=\frac{ \left( x+1 \right)\left( x+2 \right) }{ \left( x-1 \right)\left( x-2\right) }\left\{ \frac{ 1 }{x+1 }+\frac{ 1 }{x+2 }-\frac{ 1 }{x-1 }-\frac{ 1 }{x-2 } \right\}\]