Question

Differentiate each the following from first principles :

$\left(i\right)e^{-x}$

$\left(ii\right)e^{3x}$

$\left(iii\right)e^{ax+b}$

$\left(iv\right)x{e}^x$

$\left(v\right)x^2{e}^x$

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

$\left(vii\right)e^{\sqrt{2x}}$

$\left(viii\right)e^{\sqrt{ax+b}}$

$\left(ix\right)a^{\sqrt{x}}$

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

Answer 1

(i) We have,

$f\left(x\right)=e^{-x}$

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

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

$=-e^{-x}[li{m}_{h\to 0}\frac{e^{\theta -1}}{\theta }=1]$

(ii) We have,

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

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

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

Multiplying Numberator and Denominator by 3

$=li{m}_{h\to 0}{e}^{3\left(x\right)}\frac{e^{3h}-1}{3h}[\because li{m}_{h\to 0}\frac{e^{3h}-1}{3}=1]$

$=3e^{3x}$

(iii) We have,

$f\left(x\right)=e^{ax+b}$

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

$=li{m}_{h\to 0}\frac{e^{a(x+h)+b}-e^{ax+b}}{h}$

$=li{m}_{h\to 0}\frac{e^{ax}\times e^{ah}\times e^b-e^{ax}\times e^b}{h}$

$=li{m}_{h\to 0}\frac{e^b\times e^{ax}\left(e^{ah-1}\right)}{h}$

$=li{m}_{h\to 0}{e}^{ax+b}\times \frac{a(e^{ah}-1)}{a.h}$

Multiplying Numerator and denominator by a

$[\because li{m}_{h\to 0}\frac{(e^{ah}-1)}{ah}=1]$

$=a{e}^{ax+b}.$

(iv) We have,

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

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

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

$=li{m}_{h\to 0}\frac{x{e}^x.e^h+h{e}^x.e^h-x{e}^k}{h}$

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

$=x{e}^x+e^x=e^x(x+1)$

(v) We have,

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

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

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

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

$[\because \frac{e^h-1}{h}-1]$

$\therefore x^2{e}^x+e^x(0+2x)$

$=x^2{e}^x+2x{e}^x$

$=e^x(x^2+2x)$

(vi) We have,

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

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

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

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

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

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

$\because h\to 0$

$\Rightarrow 2xh+h^2=0$

$\text{and}li{m}_{0\to 0}\frac{e^{\theta }-1}{\theta }=1$

$=li{m}_{h\to 0}{e}^{x^2+1}.1\times 2x+h=2x{e}^{x^2+1}$

(vii) We have,

$f\left(x\right)=e\sqrt{2x}$

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

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

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

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

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

$\because h\to \Rightarrow \sqrt{2(x+h)}-\sqrt{2x}\Rightarrow 0$

$\text{and}li{m}_{\theta \to 0}\frac{e^0-1}{\theta }=1$

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

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

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

$=e^{\sqrt{2x}}\times \frac{1}{2\sqrt{2x}}$

(viii) We have,

$f\left(x\right)=e^{ax+b}$

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

$=li{m}_{h\to 0}\frac{e^{\sqrt{a(x+h)+b}}-e^{\sqrt{ax+b}}}{h}$

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

$=li{m}_{h\to 0}{e}^{\sqrt{ax+b}}\times \frac{e^{\sqrt{a(x+h)+b}-\sqrt{ax+b}}-1}{\sqrt{a\sqrt{(x+h)+b}-\sqrt{ax+b}}}\times \frac{\sqrt{a(x+h)+b}-\sqrt{ax+b}}{h}$

Multiplying Numerator and Denominator by

$\sqrt{a(x+h)+b}-\sqrt{ax+b}$

$\because h\to 0$

$\therefore \sqrt{a(x+h)+b}-\sqrt{ax+b}=0$

$\text{and}li{m}_{0\to 0}\frac{e^0-1}{\theta }=1$

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

Again multiplied Numerator and Denominator by

$\sqrt{a(x+h)+b}+\sqrt{ax+b}$

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

$=\frac{e^{\sqrt{ax+b}}\times a}{2\sqrt{ax+b}}$

$=\frac{a{e}^{\sqrt{ax+b}}}{2\sqrt{ax+b}}$

$\left(ix\right)a^{\sqrt{x}}$

$f\left(x\right)=a^{\sqrt{x}}=e^{\sqrt{x}loga}$

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

$=li{m}_{h\to 0}\frac{e^{\sqrt{x+h}loga}-e^{\sqrt{x}}loga}{h}$

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

Multiplying numerator and denominator by

$(\sqrt{x+h}-\sqrt{x})loga$

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

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

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

Multiply numerator and denominator by $\sqrt{x+h}-\sqrt{x}$

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

$=e^{\sqrt{x}loga}li{m}_{h\to 0}loga\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$

$=e^{\sqrt{x}}loga\frac{loga}{2\sqrt{x}}$

$=\frac{a\sqrt{x}}{2\sqrt{x}}lo{g}_{e}a$

(x) We have,

$f\left(x\right)=3^{x^2}=e^{x^{2}log3}$

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

$=li{m}_{h\to 0}\frac{e^{(x+h{)}^{2}log3}-3^{x^{2}log3}}{h}$

$=li{m}_{h\to 0}{e}^{x^{2}log3}\frac{[{(e^{(x+h{)}^2}-x^2)}^{log3}-1]}{h}$

$=li{m}_{h\to 0}{e}^{x^2}log3\frac{{[e^{(x+h{)}^2}-x^2]}^{log3}-1}{(x+h{)}^2-x^2}\times \frac{(x+h{)}^2-x^2}{h}$

Multiplying Numerator and Denominator by $(x+h{)}^2-x^2$

$\therefore li{m}_{h\to 0}{e}^{x^{2}log3}\times \frac{(x+h+x)(x+h-x)}{h}$

$=e^{x^{2}log3}\times 2x$

$=2x{e}^{x^{2}log3}$

$=2x{3}^{x^{2}log3}$