The derivative of $sin\, x$ with respect to $ cos\text{ }x $ is
- $ sin\,\,x $
- $ -cos\,x $
- $ tan\,\,x $
- $ -cot\,x $
The Correct Option is D
Solution and Explanation
and $ v(x)=\cos \,x\,\,\,\Rightarrow \,\,\frac{du}{dx}=\cos \,x $
and $ \frac{dv}{dx}=-\sin x $
Then, $ \frac{du}{dv}=\frac{du/dx}{dv/dx}\Rightarrow \frac{du}{dv}=\frac{\cos \,x}{-\sin x} $
$ \Rightarrow $ $ \frac{du}{dv}=-\cot \,x $
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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.