Question

10. Use the limit definition of the derivative to calculate the derivatives of the following functions. a. f(x) = 2x2 – 3x + 4 b. g(x) = = x2 +1 1 x2 +1 c. h(x) = 3x - 2 a. 11. Find the derivative with respect to x. x² - 4x f(x)= b. y = sec v c. 5x2 – 2xy + 7y2 = 0 1+cos x 1-cosx cos(Inu) e. S(x) = du 1+1 + + f. y =sin(x+y) g. h(x)=xVx+1 h. y = tan-'(3x)+Incosx| i xe” +1= xy j: y=xe" k. y = sinx d.y= 1. m(x)= \n(xVx+7) m. f(x)=(2x*+5) ,ות

Answer 1

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10. (a) f(x) 2 we know that, f(x) = 2x²-3x+4. unit definition of derivative is f'(x) = lim frath) - f(x) ho h / f(x) = 2x²-3x+4 f(x+b) = 2(x+h)-3(x+h) + 4 = 2 (x'+u+ 2 ks] – 3(x+h) + 4 2x²+2h+4hx - 3x-3h + 4. fath) then f'(x) = lim f(x+h)-f(x) h f'(x) = lim { (2x² + 25 + 4 h 2-3x-3h+ 4) = (2x²-3x+4); h f(x) = lim (277-25+ 4 hx - 34-3h+- 2 + 3x -4. +34-4] h - 2 4²+ 4 hx-3h [ *2f+962- Ho tim - Lim ho h (-2h +4x-3) h Lim (-2h + 4 x-3) た - 2x ot 4x-3 4X-3 Hence, If'(x) = 4x-3 (Any)

(6) gox) = x 4 g'(x) Uim ho g(x+h)-g(x) h gay - - Lim (x+h+1 x²+1 h - lim ho 1 2+1 - ((x+1)+1) [cx+h3+1] (x+1) h #+1 - [x'+h+2hX+1 ] h (ex+v'+1) (x + 1) Top stalim hto im ho = lim h(oxendat) (x+1) lim (h+24) h ((x+n'+1) (41) - (h+2x) ho = lim hto 1 (mm)+1)(2+0) -22 22 - (0+2x) (240)+1) (2+1) (x + 1) (x + 1) g'(x) = (x + 1)2 (71) 24 Hence, Any

h(x) = √3x-2 h'(x) = lim h (ata) - g(x) 070 * √ 3 (x+9) -2 - - √3x-2 Lim ato 3X+3a-2 √3x-2 - lim ato a lim √3x +39-2-√34-2 3x+39-2 + √3x-2 Х a √3x +39-2 + √3x-2 La Lim (J31+34-274(532-2)27 a to a [√3x439-2 + 5 -√3x-2) lim 3x+39-2-x-2) ca (134+39-2 + 132-2 im 358430-7-3x+2 a (√3x +39-2+ √3x-2 39 donato Lave 13x439 - 2 + √3x-2 3 3 (3x + 3X0-2 + √3x-2 √32-2 + √3x-2 3 2√3x-2 leich) Ann ato lim lim J √3x+39-2 + √3x-2 2√3x-2 Hence,