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Question Number 46907 by Raj Singh last updated on 02/Nov/18

Answered by ajfour last updated on 02/Nov/18

$$y=(x^2+3)(x^3-5)$$$$y+△y=[{(x+△x)}^2+3][{(x+△x)}^3-5]$$$$△y={(x+△x)}^5+3{(x+△x)}^3$$$$-5{(x+△x)}^2-15$$$$-x^5-3x^3+5x^2+15$$$$={(x+△x)}^5-x^5+3[{(x+△x)}^3-x^3]$$$$-5[(x+△x)-x]$$$$=(△x)[5x^4+3\times 3x^2-5\times 2x+△x(...)]$$$$\Rightarrow \underset{△x\to 0}{\lim }\frac{△y}{△x}=5x^4+9x^2-10x$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$tocheck:$$$$\frac{dy}{dx}=2x(x^3-5)+3x^2(x^2+3)$$$$=5x^4+9x^2-10x.$$

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