Respuesta :
[tex]\\ \sf\longmapsto \dfrac{d}{dx}(ax^2+bx+c)(cx+d)[/tex]
[tex]\boxed{\sf \dfrac{d}{dx}f(x).g(x)=f(x)\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}f(x)}[/tex]
- c and d are constants
[tex]\\ \sf\longmapsto (ax^2+bx+c)\dfrac{d}{dx}(cx+d)+(cx+d)\dfrac{d}{dx}(ax^2+bx+c)[/tex]
[tex]\\ \sf\longmapsto (ax^2+bx+c)(c)+(cx+d)(2ax+b)[/tex]
[tex]\\ \sf\longmapsto acx^2+bcx+c^2+2acx^2+bcx+2adx+bd[/tex]
[tex]\\ \sf\longmapsto 3acx^2+2bcx+2adx+bd+c^2[/tex]
Answer:
• Product rule is as below:
[tex]{ \boxed{ \tt{ \: \frac{dy}{dx} = { \huge{v}} \frac{du}{dx} + { \huge{u}} \frac{dv}{dx} }}} \\ [/tex]
- u is (ax² + bx + c)
- v is (cx + d)
- du/dx is 2ax + bx
- dv/dx is c
[tex] \hookrightarrow \: { \rm{ \frac{dy}{dx} = (cx + d)(2ax + b) + (ax {}^{2} + bx + c)(c) }} \\ \\ { \rm{ \frac{dy}{dx} = (2ac {x}^{2} + bcx + 2adx + db) + (ac {x}^{2} + bcx + {c}^{2} )}} \\ \\ { \boxed{ \rm{ \frac{dy}{dx} = 3ac {x}^{2} + \{2bcx + 2adx \}x + (db + {c}^{2}) }}}[/tex]
