Question

Differentiation of e^ax by ab-initio method.

Answer 1

For this question, you are required to differentiate

$y=e^{ax}$ from first principles, i.e using the ab-initio method.

For this problem, the result

$\lim_{h \to 0} \frac{e^{ah}-1}{h}=a$

will become very useful in finding the derivative. The constant h is the infinitesimal change in x used in computing the derivative.

                                              Step 1

Define $f(x)=e^{ax}$ and $f(x+h)=e^{a(x+h)}$.

                                               Step 2

Work out the value of ,

$f(x+h)-f(x) = e^{a(x+h)}-e^{ax}\\\\f(x+h)-f(x) =e^{ax}\cdot e^{ah}-e^{ax}\\\\f(x+h)-f(x)=e^{ax}(e^{ah}-1)$.

                                               Step 3

Divide the expression in step 2 by the infinitesimal change in x, which we call h in this problem,

$\frac{f(x+h)-f(x)}{h}=\frac{e^{ax}(e^{ah}-1)}{h}$.

                                                Step 4

Work out the value of the limit of the expression in step 3 to find the derivative.  

$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}= \lim_{x \to 0} \frac{e^{ax}(e^{ax}-1)}{h}= e^{ax} \lim_{x \to 0}\frac{(e^{ax}-1)}{h} =e^{ax}\cdot a=ae^{ax}$.

Answer 2

Answer:

let y=eax-(1)

Giving small increments ∆x to x and ∆y to y

y+∆y=eà(x+∆x)-(2)

Subtracting equation from 1 to 2 ,we get

y+∆y-y=eà(x+∆x)- eàx

∆y=eà(x+∆x)_eàx

∆y=eàx+a∆x-eàx

∆y=eàx×eà∆x-eàx

∆y=eàx(eàx-1)

Dividing both sides by ∆x ,we get

∆y/∆x=eàx(3à∆x-1)/∆x

Taking limits on both sides as ∆x tends to 0,we get

limit x tends to 0∆y/∆x=limit x tends to 0 eàx(eà∆x-1)/∆x

lt.∆x tends to 0(∆y/∆x=lt.x tends to 0 eàx(eà∆x-1)/a×∆x/a

lt.∆x tends to 0 aeax(eà∆x-1)/a∆x

lt.x tends to 0 aeax×lt ∆x tends to 0(eà∆x-1)/a∆x

But lt.x tends to 0(ex-1)/x=1

lt x tends to 0(∆y/∆x=lt ∆x tends to zero aeàx×1

dy /DX =lt x tends to zero aeàx

dy/DX=aeàx..

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