P r o d u c t R u l e {\displaystyle \color {Purple}Product\,Rule} d d x [ f ⋅ g ] = d d x [ f ] ⋅ g + d d x [ g ] ⋅ f {\displaystyle {\frac {d}{dx}}[f\cdot {g}]={\frac {d}{dx}}[f]\cdot {g}+{\frac {d}{dx}}[g]\cdot {f}} Q u o t i e n t R u l e {\displaystyle \color {Blue}Quotient\,Rule} d d x [ f g ] = d d x [ f ] ⋅ g − d d x [ g ] ⋅ f g 2 {\displaystyle {\frac {d}{dx}}[{\frac {f}{g}}]={\frac {{\frac {d}{dx}}[f]\cdot {g}-{\frac {d}{dx}}[g]\cdot {f}}{g^{2}}}} E x a m p l e s {\displaystyle \mathbf {\color {Green}{Examples}} } E x .1 {\displaystyle \mathbf {Ex.1} } i f f ( x ) = x ⋅ e x {\displaystyle if\,f(x)=x\cdot {e^{x}}} f ′ ( x ) = 1 ⋅ e x + x ⋅ e x {\displaystyle f^{\prime }(x)=1\cdot {e^{x}}+x\cdot {e^{x}}} E x .2 {\displaystyle \mathbf {Ex.2} } i f f ( t ) = t ( a + b t ) {\displaystyle if\,f(t)={\sqrt {t}}(a+bt)} f ′ ( t ) = 1 2 t ( a + b t ) + t t ( b ) {\displaystyle f^{\prime }(t)={\frac {1}{2{\sqrt {t}}}}(a+bt)+t{\sqrt {t}}(b)} E x .3 {\displaystyle \mathbf {Ex.3} } i f f ( x ) = x ⋅ g ( x ) {\displaystyle if\,f(x)={\sqrt {x}}\cdot {g(x)}} g ( 4 ) = 2 {\displaystyle g(4)=2} g ′ ( 4 ) = 3 {\displaystyle g^{\prime }(4)=3} f ′ ( x ) = 1 2 x ⋅ g ( x ) + x ⋅ g ′ ( x ) {\displaystyle f^{\prime }(x)={\frac {1}{2{\sqrt {x}}}}\cdot {g(x)}+{\sqrt {x}}\cdot {g^{\prime }(x)}} E x .4 {\displaystyle \mathbf {Ex.4} } y = x 2 + x − 2 x 3 + 6 {\displaystyle y={\frac {\color {Blue}{x^{2}+x-2}}{\color {Red}{x^{3}+6}}}} d d x = y ′ = ( 2 x + 1 ) ( x 3 − 6 ) − ( x 2 + x − 2 ) ( 3 x 2 ) ( x 3 + 6 ) 2 {\displaystyle {\frac {d}{dx}}=y^{\prime }={\frac {(2x+1)(x^{3}-6)-\color {Blue}{(x^{2}+x-2)}(3x^{2})}{\color {Red}{(x^{3}+6)^{2}}}}} = ( 2 x 4 + x 4 + x 3 + 12 x + 6 − [ 3 x 4 + 3 x 2 − 6 x 2 ] ( x 3 + 6 ) 2 {\displaystyle ={\frac {(2x^{4}+x^{4}+x^{3}+12x+6-[3x^{4}+3x^{2}-6x^{2}]}{(x^{3}+6)^{2}}}} = − x 4 − 2 x 3 + 6 x 2 + 12 x + 6 ( x 3 + 6 ) 2 {\displaystyle ={\frac {-x^{4}-2x^{3}+6x^{2}+12x+6}{(x^{3}+6)^{2}}}} E x .5 {\displaystyle \mathbf {Ex.5} } y = e x 1 + x 2 ( 1 , e 2 ) {\displaystyle y={\frac {e^{x}}{1+x^{2}}}\,(1,{\frac {e}{2}})\,} d d x = e x ⋅ ( 1 + x 2 ) − e x ( 2 x ) ( 1 + x 2 ) 2 {\displaystyle {\frac {d}{dx}}={\frac {e^{x}\cdot (1+x^{2})-e^{x}(2x)}{(1+x^{2})^{2}}}} d d x | x = 1 e ( 1 + 1 ) − e ′ ( 2 ) ( 1 + 1 ) 2 = 2 e − 2 e 2 2 = 0 4 = 0 {\displaystyle {\frac {d}{dx}}|_{x=1}{\frac {e(1+1)-e^{\prime }(2)}{(1+1)^{2}}}={\frac {2e-2e}{2^{2}}}={\frac {0}{4}}=0}
![{\displaystyle {\frac {d}{dx}}[f\cdot {g}]={\frac {d}{dx}}[f]\cdot {g}+{\frac {d}{dx}}[g]\cdot {f}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/57ee97cfd2a83cef00e5454484f44a5c99a73190)
![{\displaystyle {\frac {d}{dx}}[{\frac {f}{g}}]={\frac {{\frac {d}{dx}}[f]\cdot {g}-{\frac {d}{dx}}[g]\cdot {f}}{g^{2}}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/33117095dc4fc9bab8b77262a36eac752e1beaa5)
![{\displaystyle ={\frac {(2x^{4}+x^{4}+x^{3}+12x+6-[3x^{4}+3x^{2}-6x^{2}]}{(x^{3}+6)^{2}}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/fdb5247377f2d161c9bc989a350a4642c753c895)
