Find \( \frac {dy}{dx}:\) \(sin^2y+cos\ xy=\pi\)
Solution and Explanation
The given relationship is sin2y + cos xy = \(\pi\)
Differentiating this relationship with respect to x, we obtain
\(\frac {d}{dx}\)(sin2y + cos xy) = \(\frac {d}{dx}\)(\(\pi\))
⇒\(\frac {d}{dx}\)(sin2y) + \(\frac {d}{dx}\) (cos xy) = 0 ……..... (1)
Using chain rule, we obtain
\(\implies\)\(\frac {d}{dx}\)(sin2y) = 2sin y . \(\frac {d}{dx}\)(sin y) = 2sin y cos y \(\frac {dy}{dx}\) ……...... (2)
\(\frac {d}{dx}\)(cos xy) = -sin xy. \(\frac {d}{dx}\)(xy) = - sin xy [y\(\frac {d}{dx}\)(x) + x\(\frac {d}{dx}\)]
= -sin xy [y.1+x\(\frac {dy}{dx}\)] = -y sin xy - x sin xy . \(\frac {dy}{dx}\) ………..... (3)
From (1), (2) and (3), we obtain
2sin y cos y \(\frac {dy}{dx}\) - y sin xy - x sin xy \(\frac {dy}{dx}\) = 0
\(\implies\)(2sin y cos y - x sin xy)\(\frac {dy}{dx}\) = y sin xy
\(\implies\)(sin 2y - x sin xy)\(\frac {dy}{dx}\) = y sin xy
∴\(\frac {dy}{dx}\) = \(\frac {y sin \ xy}{sin\ 2y-xsin\ xy}\)
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What is the Planning Process?
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Notes on Continuity and differentiability
Concepts Used:
Differentiability
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
- Differentiable at every point of interval (a, b)
- Right derivative exists at x = a
- Left derivative exists at x = b.