Differentate eax by ab-initio method
Answers
Explanation:
For this question, you are required to differentiate
y=e^{ax}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}=alim
h→0
h
e
ah
−1
=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}f(x)=e
ax
and f(x+h)=e^{a(x+h)}f(x+h)=e
a(x+h)
.
Step 2
Work out the value of ,
\begin{gathered}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)\\\end{gathered}
f(x+h)−f(x)=e
a(x+h)
−e
ax
f(x+h)−f(x)=e
ax
⋅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}
h
f(x+h)−f(x)
=
h
e
ax
(e
ah
−1)
.
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}lim
h→0
h
f(x+h)−f(x)
=lim
x→0
h
e
ax
(e
ax
−1)
=e
ax
lim
x→0
h
(e
ax
−1)
=e
ax
⋅a=ae
ax
.
