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Addition Rule of Differentiation

i e-xii e3 xi...

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

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

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

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

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Solution

$(i) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{x}) = \lim_{h \to 0} \frac{e^{- (x + h)} - e^{- x}}{h} = \lim_{h \to 0} \frac{e^{- x} e^{- h} - e^{- x}}{h} = \lim_{h \to 0} \frac{e^{- x} (e^{- h} - 1)}{h} = - e^{- x} \lim_{h \to 0} \frac{e^{- h} - 1}{- h} = - e^{- x} (1) = - e^{- x}$

$(ii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{3 x}) = \lim_{h \to 0} \frac{e^{3 (x + h)} - e^{3 x}}{h} = \lim_{h \to 0} \frac{e^{3 x} e^{3 h} - e^{3 x}}{h} = \lim_{h \to 0} \frac{e^{3 x} (e^{3 h} - 1)}{3 h} = 3 e^{3 x} \lim_{h \to 0} \frac{e^{3 h} - 1}{3 h} = 3 e^{3 x} (1) = 3 e^{3 x}$

$(iii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{a x + b}) = \lim_{h \to 0} \frac{e^{a (x + h) + b} - e^{a x + b}}{h} = \lim_{h \to 0} \frac{e^{a x + b} e^{a h} - e^{a x + b}}{h} = \lim_{h \to 0} \frac{e^{a x + b} (e^{a h} - 1)}{h} = a e^{a x + b} \lim_{h \to 0} \frac{e^{a h} - 1}{a h} = a e^{a x + b} (1) = a e^{a x + b}$

$(iv) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (x e^{x}) = \lim_{h \to 0} \frac{(x + h) e^{(x + h)} - x e^{x}}{h} = \lim_{h \to 0} \frac{(x + h) e^{x} e^{h} - x e^{x}}{h} = \lim_{h \to 0} \frac{x e^{x} e^{h} + h e^{x} e^{h} - x e^{x}}{h} = \lim_{h \to 0} \frac{x e^{x} e^{h} - x e^{x}}{h} + \lim_{h \to 0} \frac{h e^{x} e^{h}}{h} = \lim_{h \to 0} \frac{x e^{x} (e^{h} - 1)}{h} + \lim_{h \to 0} e^{x} e^{h} = x e^{x} (1) + e^{x} (e^{0}) = x e^{x} + e^{x}$

$(v) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (x^{2} e^{x}) = \lim_{h \to 0} \frac{(x + h)^{2} e^{(x + h)} - x^{2} e^{x}}{h} = \lim_{h \to 0} \frac{(x^{2} + 2 x h + h^{2}) e^{x} e^{h} - x^{2} e^{x}}{h} = \lim_{h \to 0} \frac{x^{2} e^{x} e^{h} + 2 x h e^{x} e^{h} + h^{2} e^{x} e^{h} - x^{2} e^{x}}{h} = \lim_{h \to 0} \frac{x^{2} e^{x} e^{h} - x^{2} e^{x}}{h} + \lim_{h \to 0} \frac{2 x h e^{x} e^{h}}{h} + \lim_{h \to 0} \frac{h^{2} e^{x} e^{h}}{h} = \lim_{h \to 0} \frac{x^{2} e^{x} (e^{h} - 1)}{h} + \lim_{h \to 0} 2 x e^{x} e^{h} + \lim_{h \to 0} h e^{x} e^{h} = x^{2} e^{x} (1) + 2 x e^{x} (1) + 0 = x^{2} e^{x} + 2 x e^{x} = (x^{2} + 2 x) e^{x}$

$(vi) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{x^{2} + 1}) = \lim_{h \to 0} \frac{e^{(x + h)^{2} + 1} - e^{x^{2} + 1}}{h} = \lim_{h \to 0} \frac{e^{x^{2} + h^{2} + 2 x h + 1} - e^{x^{2} + 1}}{h} = \lim_{h \to 0} \frac{e^{x^{2} + 1} e^{h^{2} + 2 x h} - e^{x^{2} + 1}}{h} = \lim_{h \to 0} \frac{e^{x^{2} + 1} (e^{h (h + 2 x)} - 1)}{h} \times \frac{(h + 2 x)}{(h + 2 x)} = e^{x^{2} + 1} \lim_{h \to 0} \frac{e^{h (h + 2 x)} - 1}{h (h + 2 x)} \lim_{h \to 0} (h + 2 x) = e^{x^{2} + 1} (1) (2 x) = 2 x e^{x^{2} + 1}$

$(vii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{\sqrt{2 x}}) = \lim_{h \to 0} \frac{e^{\sqrt{2 (x + h)}} - e^{\sqrt{2 x}}}{h} = 2 \lim_{h \to 0} \frac{e^{\sqrt{2 x + 2 h}} - e^{\sqrt{2 x}}}{2 x + 2 h - 2 x} = 2 \lim_{h \to 0} \frac{e^{\sqrt{2 x}} (e^{\sqrt{2 x + 2 h} - \sqrt{2 x}} - 1)}{{(\sqrt{2 x + 2 h})}^{2} - {(\sqrt{2 x})}^{2}} = 2 e^{\sqrt{2 x}} \lim_{h \to 0} \frac{e^{\sqrt{2 x + 2 h} - \sqrt{2 x}} - 1}{(\sqrt{2 x + 2 h} - \sqrt{2 x}) (\sqrt{2 x + 2 h} + \sqrt{2 x})} = 2 e^{\sqrt{2 x}} \lim_{h \to 0} \frac{e^{\sqrt{2 x + 2 h} - \sqrt{2 x}} - 1}{(\sqrt{2 x + 2 h} - \sqrt{2 x})} \lim_{h \to 0} \frac{1}{(\sqrt{2 x + 2 h} + \sqrt{2 x})} = 2 e^{\sqrt{2 x}} (1) \frac{1}{2 \sqrt{2 x}} = \frac{e^{\sqrt{2 x}}}{\sqrt{2 x}}$

$(viii) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (e^{\sqrt{a x + b}}) = \lim_{h \to 0} \frac{e^{\sqrt{a x + a h + b}} - e^{\sqrt{a x + b}}}{h} = a \lim_{h \to 0} \frac{e^{\sqrt{a x + a h + b}} - e^{\sqrt{a x + b}}}{(a x + a h + b) - (a x + b)} = a \lim_{h \to 0} \frac{e^{\sqrt{a x + b}} (e^{\sqrt{a x + a h + b} - \sqrt{a x + b}} - 1)}{{(\sqrt{(a x + a h + b)})}^{2} - {(\sqrt{(a x + b)})}^{2}} = a e^{\sqrt{a x + b}} \lim_{h \to 0} \frac{e^{\sqrt{a x + a h + b} - \sqrt{a x + b}} - 1}{(\sqrt{(a x + a h + b)} - \sqrt{(a x + b)}) (\sqrt{(a x + a h + b)} + \sqrt{(a x + b)})} = a e^{\sqrt{a x + b}} \lim_{h \to 0} \frac{e^{\sqrt{a x + a h + b} - \sqrt{a x + b}} - 1}{\sqrt{(a x + a h + b)} - \sqrt{(a x + b)}} \lim_{h \to 0} \frac{1}{\sqrt{(a x + a h + b)} + \sqrt{(a x + b)}} = a e^{^{\sqrt{a x + b}}} (1) \frac{1}{2 \sqrt{a x + b}} = \frac{a e^{^{\sqrt{a x + b}}}}{2 \sqrt{a x + b}}$

$(ix) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (a^{\sqrt{x}}) = \lim_{h \to 0} \frac{a^{\sqrt{x + h}} - a^{\sqrt{x}}}{h} = \lim_{h \to 0} \frac{a^{\sqrt{x}} (a^{\sqrt{x + h} - \sqrt{x}} - 1)}{(x + h) - (x)} = a^{\sqrt{x}} \lim_{h \to 0} \frac{(a^{\sqrt{x + h} - \sqrt{x}} - 1)}{{(\sqrt{x + h})}^{2} - {(\sqrt{x})}^{2}} = a^{\sqrt{x}} \lim_{h \to 0} \frac{(a^{\sqrt{x + h} - \sqrt{x}} - 1)}{(\sqrt{x + h} - \sqrt{x}) (\sqrt{x + h} + \sqrt{x})} = a^{\sqrt{x}} \lim_{h \to 0} \frac{(a^{\sqrt{x + h} - \sqrt{x}} - 1)}{(\sqrt{x + h} - \sqrt{x})} \lim_{h \to 0} \frac{1}{(\sqrt{x + h} + \sqrt{x})} = a^{\sqrt{x}} \log_{e} a \frac{1}{2 \sqrt{x}} = \frac{1}{2 \sqrt{x}} a^{\sqrt{x}} \log_{e} a$

$(x) \frac{d}{d x} (f (x)) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} \frac{d}{d x} (3^{x^{2}}) = \lim_{h \to 0} \frac{3^{{(x + h)}^{2}} - 3^{x^{2}}}{h} = \lim_{h \to 0} \frac{3^{x^{2} + 2 x h + h^{2}} - 3^{x^{2}}}{h} = \lim_{h \to 0} \frac{3^{x^{2}} (3^{x^{2} + 2 x h + h^{2} - x^{2}} - 1)}{h} \times \frac{(h + 2 x)}{(h + 2 x)} = 3^{x^{2}} \lim_{h \to 0} \frac{3^{h (h + 2 x)} - 1}{h (h + 2 x)} \lim_{h \to 0} (h + 2 x) = 3^{x^{2}} \log 3 (2 x) = 2 x 3^{x^{2}} \log 3$

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Similar questions

Differentiate each the following from first principles :

$(i) e^{- x}$

$(i i) e^{3 x}$

$(i i i) e^{a x + b}$

$(i v) x e^{x}$

$(v) x^{2} e^{x}$

$(v i) e^{x^{2} + 1}$

$(v i i) e^{\sqrt{2 x}}$

$(v i i i) e^{\sqrt{a x + b}}$

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

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

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(ii) e^3x

(iii) e^ax^{+ b}

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y = e^{a x} - e^{- a x}

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\infty

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