Question Number 46907 by Raj Singh last updated on 02/Nov/18
Answered by ajfour last updated on 02/Nov/18
![y=(x^2 +3)(x^3 −5) y+△y=[(x+△x)^2 +3][(x+△x)^3 −5] △y=(x+△x)^5 +3(x+△x)^3 −5(x+△x)^2 −15 −x^5 −3x^3 +5x^2 +15 =(x+△x)^5 −x^5 +3[(x+△x)^3 −x^3 ] −5[(x+△x)−x] = (△x)[5x^4 +3×3x^2 −5×2x+△x(...)] ⇒ lim_(△x→0) ((△y)/(△x)) = 5x^4 +9x^2 −10x ________________________ to check: (dy/dx)=2x(x^3 −5)+3x^2 (x^2 +3) =5x^4 +9x^2 −10x .](https://www.tinkutara.com/question/Q46911.png)
$${y}=\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left({x}^{\mathrm{3}} −\mathrm{5}\right) \\ $$$${y}+\bigtriangleup{y}=\left[\left({x}+\bigtriangleup{x}\right)^{\mathrm{2}} +\mathrm{3}\right]\left[\left({x}+\bigtriangleup{x}\right)^{\mathrm{3}} −\mathrm{5}\right] \\ $$$$\bigtriangleup{y}=\left({x}+\bigtriangleup{x}\right)^{\mathrm{5}} +\mathrm{3}\left({x}+\bigtriangleup{x}\right)^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:−\mathrm{5}\left({x}+\bigtriangleup{x}\right)^{\mathrm{2}} −\mathrm{15} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−{x}^{\mathrm{5}} −\mathrm{3}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{15} \\ $$$$\:\:\:\:\:=\left({x}+\bigtriangleup{x}\right)^{\mathrm{5}} −{x}^{\mathrm{5}} +\mathrm{3}\left[\left({x}+\bigtriangleup{x}\right)^{\mathrm{3}} −{x}^{\mathrm{3}} \right] \\ $$$$\:\:\:\:−\mathrm{5}\left[\left({x}+\bigtriangleup{x}\right)−{x}\right] \\ $$$$\:\:\:\:=\:\left(\bigtriangleup{x}\right)\left[\mathrm{5}{x}^{\mathrm{4}} +\mathrm{3}×\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}×\mathrm{2}{x}+\bigtriangleup{x}\left(…\right)\right] \\ $$$$\Rightarrow\:\:\underset{\bigtriangleup{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\bigtriangleup{y}}{\bigtriangleup{x}}\:=\:\mathrm{5}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{2}} −\mathrm{10}{x} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${to}\:{check}: \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{x}\left({x}^{\mathrm{3}} −\mathrm{5}\right)+\mathrm{3}{x}^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:=\mathrm{5}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{2}} −\mathrm{10}{x}\:. \\ $$