Question

(i) e^−x

(ii) e^3x

(iii) e^{ax + b}

(iv) x e^x

(v) x² e^x

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

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

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

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

(x) $3^{x^2}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(e^x\right)=\underset{h\to 0}{\lim }\frac{e^{-(x+h)}-e^{-x}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{-x}{e}^{-h}-e^{-x}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{-x}\left(e^{-h}-1\right)}{h} \\ =-e^{-x}\underset{h\to 0}{\lim }\frac{e^{-h}-1}{-h} \\ =-e^{-x}\left(1\right) \\ =-e^{-x} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(e^{3x}\right)=\underset{h\to 0}{\lim }\frac{e^{3(x+h)}-e^{3x}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{3x}{e}^{3h}-e^{3x}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{3x}\left(e^{3h}-1\right)}{3h} \\ =3e^{3x}\underset{h\to 0}{\lim }\frac{e^{3h}-1}{3h} \\ =3e^{3x}\left(1\right) \\ =3e^{3x} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(e^{ax+b}\right)=\underset{h\to 0}{\lim }\frac{e^{a(x+h)+b}-e^{ax+b}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{ax+b}{e}^{ah}-e^{ax+b}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{ax+b}\left(e^{ah}-1\right)}{h} \\ =a{e}^{ax+b}\underset{h\to 0}{\lim }\frac{e^{ah}-1}{ah} \\ =a{e}^{ax+b}\left(1\right) \\ =a{e}^{ax+b} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(x{e}^x\right)=\underset{h\to 0}{\lim }\frac{(x+h)e^{(x+h)}-x{e}^x}{h} \\ =\underset{h\to 0}{\lim }\frac{(x+h)e^x{e}^h-x{e}^x}{h} \\ =\underset{h\to 0}{\lim }\frac{x{e}^x{e}^h+h{e}^x{e}^h-x{e}^x}{h} \\ =\underset{h\to 0}{\lim }\frac{x{e}^x{e}^h-x{e}^x}{h}+\underset{h\to 0}{\lim }\frac{h{e}^x{e}^h}{h} \\ =\underset{h\to 0}{\lim }\frac{x{e}^x\left(e^h-1\right)}{h}+\underset{h\to 0}{\lim }{e}^x{e}^h \\ =x{e}^x\left(1\right)+e^x\left(e^0\right) \\ =x{e}^x+e^x \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(x^2{e}^x\right)=\underset{h\to 0}{\lim }\frac{(x+h{)}^2{e}^{(x+h)}-x^2{e}^x}{h} \\ =\underset{h\to 0}{\lim }\frac{(x^2+2xh+h^2)e^x{e}^h-x^2{e}^x}{h} \\ =\underset{h\to 0}{\lim }\frac{x^2{e}^x{e}^h+2xh{e}^x{e}^h+h^2{e}^x{e}^h-x^2{e}^x}{h} \\ =\underset{h\to 0}{\lim }\frac{x^2{e}^x{e}^h-x^2{e}^x}{h}+\underset{h\to 0}{\lim }\frac{2xh{e}^x{e}^h}{h}+\underset{h\to 0}{\lim }\frac{h^2{e}^x{e}^h}{h} \\ =\underset{h\to 0}{\lim }\frac{x^2{e}^x\left(e^h-1\right)}{h}+\underset{h\to 0}{\lim }2x{e}^x{e}^h+\underset{h\to 0}{\lim }h{e}^x{e}^h \\ =x^2{e}^x\left(1\right)+2x{e}^x\left(1\right)+0 \\ =x^2{e}^x+2x{e}^x \\ =\left(x^2+2x\right)e^x \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(e^{x^2+1}\right)=\underset{h\to 0}{\lim }\frac{e^{(x+h{)}^2+1}-e^{x^2+1}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{x^2+h^2+2xh+1}-e^{x^2+1}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{x^2+1}{e}^{h^2+2xh}-e^{x^2+1}}{h} \\ =\underset{h\to 0}{\lim }\frac{e^{x^2+1}\left(e^{h\left(h+2x\right)}-1\right)}{h}\times \frac{\left(h+2x\right)}{\left(h+2x\right)} \\ =e^{x^2+1}\underset{h\to 0}{\lim }\frac{e^{h\left(h+2x\right)}-1}{h\left(h+2x\right)}\underset{h\to 0}{\lim }\left(h+2x\right) \\ =e^{x^2+1}\left(1\right)\left(2x\right) \\ =2x{e}^{x^2+1} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(e^{\sqrt{2x}}\right)=\underset{h\to 0}{\lim }\frac{e^{\sqrt{2(x+h)}}-e^{\sqrt{2x}}}{h} \\ =2\underset{h\to 0}{\lim }\frac{e^{\sqrt{2x+2h}}-e^{\sqrt{2x}}}{2x+2h-2x} \\ =2\underset{h\to 0}{\lim }\frac{e^{\sqrt{2x}}\left(e^{\sqrt{2x+2h}-\sqrt{2x}}-1\right)}{{\left(\sqrt{2x+2h}\right)}^2-{\left(\sqrt{2x}\right)}^2} \\ =2e^{\sqrt{2x}}\underset{h\to 0}{\lim }\frac{e^{\sqrt{2x+2h}-\sqrt{2x}}-1}{\left(\sqrt{2x+2h}-\sqrt{2x}\right)\left(\sqrt{2x+2h}+\sqrt{2x}\right)} \\ =2e^{\sqrt{2x}}\underset{h\to 0}{\lim }\frac{e^{\sqrt{2x+2h}-\sqrt{2x}}-1}{\left(\sqrt{2x+2h}-\sqrt{2x}\right)}\underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{2x+2h}+\sqrt{2x}\right)} \\ =2e^{\sqrt{2x}}\left(1\right)\frac{1}{2\sqrt{2x}} \\ =\frac{e^{\sqrt{2x}}}{\sqrt{2x}} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{viii}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(e^{\sqrt{ax+b}}\right)=\underset{h\to 0}{\lim }\frac{e^{\sqrt{ax+ah+b}}-e^{\sqrt{ax+b}}}{h} \\ =a\underset{h\to 0}{\lim }\frac{e^{\sqrt{ax+ah+b}}-e^{\sqrt{ax+b}}}{\left(ax+ah+b\right)-\left(ax+b\right)} \\ =a\underset{h\to 0}{\lim }\frac{e^{\sqrt{ax+b}}\left(e^{\sqrt{ax+ah+b}-\sqrt{ax+b}}-1\right)}{{\left(\sqrt{\left(ax+ah+b\right)}\right)}^2-{\left(\sqrt{\left(ax+b\right)}\right)}^2} \\ =a{e}^{\sqrt{ax+b}}\underset{h\to 0}{\lim }\frac{e^{\sqrt{ax+ah+b}-\sqrt{ax+b}}-1}{\left(\sqrt{\left(ax+ah+b\right)}-\sqrt{\left(ax+b\right)}\right)\left(\sqrt{\left(ax+ah+b\right)}+\sqrt{\left(ax+b\right)}\right)} \\ =a{e}^{\sqrt{ax+b}}\underset{h\to 0}{\lim }\frac{e^{\sqrt{ax+ah+b}-\sqrt{ax+b}}-1}{\sqrt{\left(ax+ah+b\right)}-\sqrt{\left(ax+b\right)}}\underset{h\to 0}{\lim }\frac{1}{\sqrt{\left(ax+ah+b\right)}+\sqrt{\left(ax+b\right)}} \\ =a{e}^{{}^{\sqrt{ax+b}}}\left(1\right)\frac{1}{2\sqrt{ax+b}} \\ =\frac{a{e}^{{}^{\sqrt{ax+b}}}}{2\sqrt{ax+b}} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ix}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(a^{\sqrt{x}}\right)=\underset{h\to 0}{\lim }\frac{a^{\sqrt{x+h}}-a^{\sqrt{x}}}{h} \\ =\underset{h\to 0}{\lim }\frac{a^{\sqrt{x}}\left(a^{\sqrt{x+h}-\sqrt{x}}-1\right)}{\left(x+h\right)-\left(x\right)} \\ =a^{\sqrt{x}}\underset{h\to 0}{\lim }\frac{\left(a^{\sqrt{x+h}-\sqrt{x}}-1\right)}{{\left(\sqrt{x+h}\right)}^2-{\left(\sqrt{x}\right)}^2} \\ =a^{\sqrt{x}}\underset{h\to 0}{\lim }\frac{\left(a^{\sqrt{x+h}-\sqrt{x}}-1\right)}{\left(\sqrt{x+h}-\sqrt{x}\right)\left(\sqrt{x+h}+\sqrt{x}\right)} \\ =a^{\sqrt{x}}\underset{h\to 0}{\lim }\frac{\left(a^{\sqrt{x+h}-\sqrt{x}}-1\right)}{\left(\sqrt{x+h}-\sqrt{x}\right)}\underset{h\to 0}{\lim }\frac{1}{\left(\sqrt{x+h}+\sqrt{x}\right)} \\ =a^{\sqrt{x}}{\log }_{e}a\frac{1}{2\sqrt{x}} \\ =\frac{1}{2\sqrt{x}}{a}^{\sqrt{x}}{\log }_{e}a \end{gathered}$$

$$\begin{gathered} \left(\mathrm{x}\right)\frac{d}{dx}\left(f\left(x\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h} \\ \frac{d}{dx}\left(3^{x^2}\right)=\underset{h\to 0}{\lim }\frac{3^{{\left(x+h\right)}^2}-3^{x^2}}{h} \\ =\underset{h\to 0}{\lim }\frac{3^{x^2+2xh+h^2}-3^{x^2}}{h} \\ =\underset{h\to 0}{\lim }\frac{3^{x^2}\left(3^{x^2+2xh+h^2-x^2}-1\right)}{h}\times \frac{\left(h+2x\right)}{\left(h+2x\right)} \\ =3^{x^2}\underset{h\to 0}{\lim }\frac{3^{h\left(h+2x\right)}-1}{h\left(h+2x\right)}\underset{h\to 0}{\lim }\left(h+2x\right) \\ =3^{x^2}\log 3\left(2x\right) \\ =2x{3}^{x^2}\log 3 \end{gathered}$$