The given relationship is y = tan-1\(( \frac {3x-x^3}{1-3x^2})\)
y = tan-1\(( \frac {3x-x^3}{1-3x^2})\)
⇒tan y = \(( \frac {3x-x^3}{1-3x^2})\) …...….. (1)
It is known that, tan y = \(\frac {3tan\ \frac y3-tan^3\frac y3}{1-3tan^2\frac y3}\) …….... (2)
Comparing equations (1) and (2), we obtain
x = tan \(\frac y3\)
Differentiating this relationship with respect to x, we obtain
\(\frac {d}{dx}\)(x) = \(\frac {d}{dx}\)(tan \(\frac y3\))
⇒1 = sec2\(\frac y3\) . \(\frac {d}{dx}\)(\(\frac y3\))
⇒1 = sec2\(\frac y3\) . \(\frac 13\) . \(\frac {dy}{dx}\)
⇒\(\frac {dy}{dx}\) = \(\frac {3}{sec^2\frac y3}\) = \(\frac {3}{tan^2\frac y3}\)
∴\(\frac {dy}{dx}\)= \(\frac {3}{1+x^2}\)
What is the Planning Process?
Differentiability of a function A function f(x) is said to be differentiable at a point of its domain if it has a finite derivative at that point. Thus f(x) is differentiable at x = a
\(\frac{d y}{d x}=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\)
⇒ f'(a – 0) = f'(a + 0)
⇒ left-hand derivative = right-hand derivative.
Thus function f is said to be differentiable if left hand derivative & right hand derivative both exist finitely and are equal.
If f(x) is differentiable then its graph must be smooth i.e. there should be no break or corner.
Note:
(i) Every differentiable function is necessarily continuous but every continuous function is not necessarily differentiable i.e. Differentiability ⇒ continuity but continuity ⇏ differentiability
(ii) For any curve y = f(x), if at any point \(\frac{d y}{d x}\) = 0 or does not exist then, the point is called “critical point”.
3. Differentiability in an interval
(a) A function fx) is said to be differentiable in an open interval (a, b), if it is differentiable at every point of the interval.
(b) A function f(x) is differentiable in a closed interval [a, b] if it is