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We know the derivative of ex, which is ex.
(ex)' = ex
We can find the derivative of eln(βx) using chain rule.
Find α΅ΚΈβdβ, if
y = eln(βx)
Let u = βx.
y = eln(u)
Let v = ln(u).
y = ev
Now,
y = ev ----> y is a function of v
v = ln(u) ----> v is is a function of u
u = βx ----> u is is a function of x
By chain rule, the derivative of y with respect to x :
Substitute y = ev, v = cotu and u = βx.
Substitute v = ln(u).
Substitute u = βx.
Therefore,
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