Question

Differentiate each the following from first principles :

$\left(i\right)\frac{2}{x}$

$\left(ii\right)\frac{1}{\sqrt{x}}$

$\left(iii\right)\frac{1}{x^3}$

$\left(iv\right)\frac{x^2+1}{x}$

$\left(v\right)\frac{x^2-1}{x}$

$\left(vi\right)\frac{x+1}{x+2}$

$\left(vii\right)\frac{x+2}{3x+5}$

$\left(viii\right)k{x}^n$

$\left(ix\right)\frac{1}{\sqrt{3-x}}$

$\left(x\right)x^2+x+3$

$\left(xi\right)(x+2{)}^3$

$\left(xii\right)x^3+4x^2+3x+2$

$\left(xiii\right)(x^2+1)(x-5)$

$\left(xiv\right)\sqrt{2x^2+1}$

$\left(xv\right)\frac{2x+3}{x-2}$

Answer 1

(i) We have,

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{f\frac{2}{x+h}-\frac{2}{x}}{h}$

$=li{m}_{h\to 0}\frac{(2x-2x-2h)}{hx(x+h)}$

$=li{m}_{h\to 0}\frac{2(x-x-h)}{hx(x+h)}$

$=li{m}_{h\to 0}\frac{-2h}{h.x(x+h)}$

$=li{m}_{h\to 0}\frac{-2}{x(x+h)}$

$=\frac{-2}{x^2}$

(ii) We have,

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{\frac{1}{\sqrt{x+y}}-\frac{1}{\sqrt{x}}}{h}$

$=li{m}_{h\to 0}\frac{\frac{1}{\sqrt{x+y}}-\frac{1}{\sqrt{x}}}{h}$

$=li{m}_{h\to 0}\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h.h}}\times \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h.h}}$

$=li{m}_{h\to 0}\frac{x-(x+h)}{\sqrt{x}\sqrt{x+h.h}(\sqrt{x}+\sqrt{x+h})}$

$=li{m}_{h\to 0}\frac{-h}{\sqrt{x}\sqrt{x+h.h}(\sqrt{x}+\sqrt{x+h})}$

$=li{m}_{h\to 0}\frac{-1}{x\sqrt{x+y}+\sqrt{x}\left(\sqrt{x+h}\right)}$

$=li{m}_{h\to 0}\frac{-1}{2x\sqrt{x}}$

$=\frac{-1}{2}{x}^{-\frac{3}{2}}$

(iii) We have,

$f\left(x\right)\frac{1}{x^3}$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{\frac{1}{(x+h{)}^3}-\frac{1}{x^3}}{h}$

$=li{m}_{h\to 0}\frac{x^3-(x+h{)}^3}{x^{3}h(x+h{)}^3}$

$=li{m}_{h\to 0}\frac{x^3-(x^3+3x^{2}h+3x{h}^2+h^3)}{x^{3}h(x+h{)}^3}$

$=li{m}_{h\to 0}\frac{x^3-x^3-3x^2-3xh-h^2}{x^{3}h(x+h{)}^3}$

$=li{m}_{h\to 0}\frac{-3x^2-3xh-h^2}{x^3(x+h{)}^3}$

$=\frac{-3x^2}{x^6}=\frac{-3}{4}$

(iv) We have,

$f\left(x\right)=\frac{x^2+1}{x}$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{\frac{(x+h{)}^2+1}{(x+h)}-\frac{x^2+1}{x}}{h}$

$=li{m}_{h\to 0}\frac{x[x^2+h^2+2xh+1]-(x^2+1)(x+h)}{hx(x+h)}$

$=li{m}_{h\to 0}\frac{x^3+x{h}^2+2x^{2}h+x-x^3-x-x^{2}h-h}{hx(x+h)}$

$=li{m}_{h\to 0}\frac{xh+2x^2-x^2-1}{x(x+h)}$

$=\frac{x^2-1}{x^2}=1-\frac{1}{x^2}$

(v) We have,

$f\left(x\right)=\frac{x^2-1}{x}$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{\frac{(x+h{)}^2-1}{(x+h)}-\frac{x^2-1}{x}}{h}$

$=li{m}_{h\to 0}\frac{x(x^2+h^2+2xh-1)-(x+h)(x^2-1)}{x(x+h)h}$

$=li{m}_{h\to 0}\frac{xh+2x^2-x^2+1}{x(x+h)}$

$=\frac{x^2+1}{x^2}=1+\frac{1}{x^2}$

(vi) We have,

$f\left(x\right)=\frac{x+1}{x+2}$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{\frac{(x+h)+1}{(x+h)+2}-\frac{x+1}{x+2}}{h}=li{m}_{h\to 0}\frac{(x+2)(x+h+1)-(x+1)(x+h+2)}{(x+h+2)(x+2)h}$

$=li{m}_{h\to 0}\frac{(x^2+2x+xh+2h+2+x)-(x^2+xh+2x+x+h+2)}{(x+h+2)(x+2)h}$

$=li{m}_{h\to 0}\frac{h}{(x+h+2)(x+2)h}=\frac{1}{(x+2{)}^2}$

(vii) We have $\frac{x+2}{3x+5}$

$f\left(x\right)=\frac{x+2}{3x+5}$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{\frac{(x+h+2)}{3(x+h)+5}-\frac{x+2}{3x+5}}{h}=li{m}_{h\to 0}\frac{(3x+5)(x+h+2)-(x+2)(3x+3h+5)}{(3x+5)(3x+3h+5)h}$

$=li{m}_{h\to 0}\frac{(3x^2+5x+3xh+5h+6x+10)-(3x^2+3xh+5x+6x+6h+10)}{(3x+5)(3x+3h+5)h}$

$=li{m}_{h\to 0}\frac{-h}{(3x+h)(3x+3h+5)h}=li{m}_{h\to 0}\frac{-1}{(3x+h)(3x+3h+5)}=\frac{-1}{(3x+5{)}^2}$

(viii) We have,

$f\left(x\right)=k{x}^n$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{k(x+h{)}^n-k{x}^n}{h}$

$=kli{m}_{h\to 0}\frac{(x^n+n{x}^{n-1}h+\frac{(n-1)}{2}{x}^{n-2}{h}^2+....)-x^n}{h}=[\because (x+y{)}^n=x^n+n{x}^{n-1}y...]$

$=kli{m}_{h\to 0}n{x}^{n-1}+\frac{n(n-1)}{2!}{x}^{n-2}h+\frac{n(n-1)(n-2)}{3!}{x}^{n-3}{h}^2....$

$=kn{x}^{n-1}+0+\mathrm{0....}=kn{x}^{n-1}$

(ix) We have,

$f\left(x\right)=\frac{1}{\sqrt{3-x}}$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{\frac{1}{\sqrt{3-(x+h)}}-\frac{1}{\sqrt{3-x}}}{h}=li{m}_{h\to 0}\frac{\sqrt{3-x}-\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)\times h}}[\text{Rationalising the numerator by}\sqrt{3-x}+\sqrt{3-(x+h)1}]$

$=li{m}_{h\to 0}\frac{\sqrt{3-x}-\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)h}}\times \frac{\sqrt{3-1}+\sqrt{3-(x+h)}}{\sqrt{3-x}+\sqrt{3-(x+h)}}$

$=li{m}_{h\to 0}\frac{(3-x)-(3-(x+h\left)\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\times h\left(\sqrt{3-x}\right)+\sqrt{3-(x+h)}}$

$=li{m}_{h\to 0}\frac{h}{\sqrt{3-x}\sqrt{3-(x+h)}\times h(\sqrt{3-x}+\sqrt{3-(x+h)})}$

$=\frac{1}{(3-x)\times 2\sqrt{3-x}}=\frac{1}{2(3-x{)}^{\frac{3}{2}}}$

(x) We have,

$f\left(x\right)=x^2+x+3$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{\left\{\right(x+h{)}^2+(x+h)+3\}-x^2+x+3}{h}$

$=li{m}_{h\to 0}\frac{x^2+h^2+2xh+x+h+3-x^2-x-3}{h}$

$=li{m}_{h\to 0}\frac{2xh+h^2+h}{h}=li{m}_{h\to 0}\frac{h(2x+h+1)}{h}=2x+0+1=2x+1$

(xi) We have,

$f\left(x\right)=(x+2{)}^3$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{(x+h+2{)}^3-(x+2{)}^3}{h}=li{m}_{h\to 0}\frac{\left\{\right(x+2)+h{\}}^3-(x+2{)}^3}{h}$

$=li{m}_{h\to 0}\frac{(x+2{)}^3+h^3+3h(x+2{)}^2+3(x+2)h^2-(x+2{)}^3}{h}$

$=li{m}_{h\to 0}3(x+2{)}^3-3(x+2)h+h^2=3(x+2{)}^2$

(xii) We have,

$f\left(x\right)=x^3+4x^2+3x+2$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{(x+h{)}^3+4(x+h{)}^2+3(x+h)+2-(x^3+4x^2+3x+2)}{h}$

$\text{On solving we get,}$

$=li{m}_{h\to 0}\frac{x^3+h^3+3x^{2}h+3h^{2}x+4x^2+4h^2+8hx+3x+3h+2-x^3-4x^2-3x-2}{h}$

$=li{m}_{h\to 0}\frac{3x^{2}h+3x{h}^2+h^3+4h^2+8hx+3h}{h}$

$=li{m}_{h\to 0}3x^2+3xh+h^2+4h+8x+3=3x^2+8x+3$

(xiii) We have,

$f\left(x\right)=x^3-5x^2+x-5$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{\left\{\right(x+h{)}^3+(x+h)-5(x+h{)}^2-5\}-(x^3-5x^2+x-5)}{h}$

$=li{m}_{h\to 0}\frac{(x^3+h^3+3x^{2}h+3h^{2}x+x+h-5x^2-5h^2-10xh-5)-(x^3-5x^2+x-5)}{h}$

$=li{m}_{h\to 0}\frac{(3x^{2}h+3h^{2}x+h^3+h-5h^2-10xh)}{h}$

$=li{m}_{h\to 0}3x^2+3xh+h^2+1-5h-10x=3x^2-10x+1$

(xiv) We have,

$f\left(x\right)=\sqrt{2x^2+1}$

$\because f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}$

$=li{m}_{h\to 0}\frac{\sqrt{2(x+h{)}^2+1}-\sqrt{2x^2+1}}{h}$

Multiplying Numerator and Denominator by $\sqrt{2(x+h{)}^2+1}+\sqrt{2x^2+1}$

$=li{m}_{h\to 0}\frac{\left\{2\right(x+h{)}^2+1-(2x^2+1)\}}{h\left(\sqrt{2(x+h{)}^2}\right)+\sqrt{2x^2+1}}=li{m}_{h\to 0}\frac{2x^2+2h^2+4xh+1-2x^2-1}{h(\sqrt{2(x+h{)}^2+1}+\sqrt{2x^2+1})}$

$=li{m}_{h\to 0}\frac{4xh+2h^2}{(\sqrt{2(x+h{)}^2+1}+\sqrt{2x^2+1})}=\frac{4x}{2\sqrt{2x^2+1}}$

$=\frac{2x}{\sqrt{2x^2+1}}$

(xv) We have, $f\left(x\right)=\frac{2x+3}{x-2}$

Therefore,

$f^{\prime }\left(x\right)=li{m}_{h\to 0}\frac{f(a+h)-f\left(a\right)}{h}$

$=li{m}_{h\to 0}\frac{\left(\frac{2x+2h+3}{x+h-2}\right)-\left(\frac{2x+3}{x-2}\right)}{h}$

$=li{m}_{h\to 0}\frac{2x^2+2hx+3x-4x-4h-6-2x^2-2hx+4x-3x-3h+6}{h(x+h-2)(x-2)}$

$=li{m}_{h\to 0}\frac{-7}{(x+h-2)(x-2)}=\frac{-7}{(x-2{)}^2}$