 # Differentiability

• A function f(x) is differentiable at a point x = a if  \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} exists finitely.
• A function f(x) is differentiable at a point x = a if LHD = RHD
\Rightarrow \lim_{x \rightarrow a^{-}}\frac{f(x)-f(a)}{x-a}=\lim_{x \rightarrow a^{+}}\frac{f(x)-f(a)}{x-a}
\Rightarrow \lim_{h \rightarrow 0}\frac{f(a-h)-f(a)}{-h}=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}
⇒  f'(a) = f'(a+)
• The graph of the curve is said to be differentiable through out the domain, the graph has no sharp edges.
• Every differentiable function is continuous, but the converse is not true.
• Chain rule is rule to differentiate composites of functions. If f = v o u, t = u(x) and if both \frac{dt}{dx}\ and\ \frac{dv}{dt} exist then
\frac{df}{dx}= \frac{dv}{dt}\cdot \frac{dt}{dx}
• Derivative of Inverse trigonometric functions:
\frac{d}{dx}(\sin^{-1} \ x)=\frac{1}{\sqrt{1-x^{2}}}
\frac{d}{dx}(\cos^{-1} \ x)=\frac{-1}{\sqrt{1-x^{2}}}
\frac{d}{dx}(\tan^{-1} \ x)=\frac{1}{1+x^{2}}
\frac{d}{dx}(\cot^{-1} \ x)=\frac{-1}{1+x^{2}}
\frac{d}{dx}(\sec^{-1} \ x)=\frac{1}{|x|\sqrt{x^{2}-1}}
\frac{d}{dx}(cosec^{-1} \ x)=\frac{-1}{|x|\sqrt{x^{2}-1}}
• Fundamental Rules:
\frac{d}{dx}(k)=0
\frac{d}{dx}(u\pm v)=\frac{du}{dx}\pm \frac{dv}{dx}
\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}
\frac{d}{dx}(uvw)=\frac{du}{dx}(vw)+u\left(\frac{dv}{dx}\right)w+uv\left(\frac{dw}{dx}\right)
\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\left(\frac{dv}{dx}\right)}{v^{2}}
• Tricks on derivatives:

Tricks on continuity and differentiablity:

• Every differential function is continuous but every continuous function need not be differentiable.
Differentiable    ⇒   Continuous
• A function f(x) is not continuous is not a differentiable function
Not continuous   ⇒     Not differentiable
• A function f(x) is not differentiable but both LHD and RHD finite and not equal, then f(x) is a continuous function.
Not differentiable    →   Continuous
Finite LHD ≠ Finite RHD
• Every modulus function is continuous but not differentiable at its critical points. i.e., f(x) = |x − a| + |x − b| is continuous but not differentiable at x = a and x = b.

### Part6: View the Topic in this video From 00:18 To 14:08

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1. Differentiation of Composite Function (Chain Rule): If f and g are differentiable functions in their domain, then fog is also differentiable and (fog)' (x) = f' {g(x)} g'(x)
More easily, if y = f(u) and u = g(x), then \frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}.
2. \frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^{2}}}, -1 < x < 1
3. \frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^{2}}}, -1 < x < 1
4. \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^{2}}
5. \frac{d}{dx}(\cot^{-1}x)=-\frac{1}{1+x^{2}}
6. \frac{d}{dx}(\sec^{-1}x)=\frac{1}{|x|\sqrt{x^{2}-1}}, |x| > 1
7. \frac{d}{dx}(cosec^{-1}x)=-\frac{1}{|x|\sqrt{x^{2}-1}}, |x| > 1
8. \frac{d}{dx}(\sin h \ x)=\cos h \ x
9. \frac{d}{dx}(\cos h \ x)=\sin h \ x
10. \frac{d}{dx}(\tan h \ x)=\sec h^2 \ x
11. \frac{d}{dx}(\cot h \ x)=-cosec h^2 \ x
12. \frac{d}{dx}(\sec h \ x)=-\sec h \ x \tan h \ x
13. \frac{d}{dx}(cosec h \ x)=-cosec h \ x \cot h \ x
14. 2 \sin^{-1}x = \sin^{-1}\left(2x\sqrt{1-x^{2}}\right)
15. 2 \cos^{-1}x = \cos^{-1}(2x^{2}-1) \ or \ \cos^{-1}(1-2x^{2})
16. 2 \tan^{-1}x = \begin{cases}\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\\ \tan^{-1}\left(\frac{2x-x^{2}}{1}\right)\\ \cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\end{cases}
17. 3 \sin^{-1}x =\sin^{-1}(3x-4x^3)
18. 3 \cos^{-1}x =\cos^{-1}(4x^{3}-3x)
19. 3 \tan^{-1}x =\tan^{-1}\left(\frac{3x-x^{3}}{1-3x^{2}}\right)
20. \cos^{-1}x+\sin^{-1}x=\frac{\pi}{2}
21. \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}
22. \sec^{-1}x+cosec^{-1}x=\frac{\pi}{2}
23. \sin^{-1}x\pm \sin^{-1}y=\sin^{-1}\left[x\sqrt{1-y^{2}} \pm y\sqrt{1-x^{2}}\right]
24. \cos^{-1}x\pm \cos^{-1}y=\cos^{-1}\left[xy\mp\sqrt{(1-x^{2})(1-y^{2})}\right]
25. \tan^{-1}x\pm \tan^{-1}y=\tan^{-1}\left[\frac{x \pm y}{1 \mp xy}\right]
26. Differentiation of Parametric Functions : If x = f(t), y = g(t), where t is parameter, then
\frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}=\frac{\frac{d}{dt} \ g(t)}{\frac{d}{dt} \ f(t)}=\frac{g'(t)}{f'(t)}