Differentiability
 A function f(x) is differentiable at a point x = c, if its left hand derivative as well as right hand derivative exists finitely and are equal.
 The derivative of the function y = f(x) at any point x = c represents the slope of the tangent to the curve. So, slope of the tangent to the curve y = f(x) is represented as tan θ, which is the derivative of the function at x = c. Hence, the derivative of the function f at the point x = c will exist iff there is one and only one tangent to the curve y = f(x) at the point (c, f(c)) and is not parallel to yaxis.
 A function is said to be differentiable if at any particular point, it provides the slope of the function and there is no sharp point and breaks in the graph of the function.
 Continuity is the necessary condition for differentiability.
 A function f is said to be differentiable in an open interval (a,b) iff it is differentiable at every point of the interval (a,b).
 A function f is called differentiable in the closed interval [a,b], iff it is differentiable in (a,b) and also differentiable at a from right and at b from left.
 Derivative of trigonometric functions can be determined using first principle.
Keywords: Geometrical representation of derivative, Derivatives of Trigonometric Functions, Derivatives of Polynomial Functions
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Q1Marks:1
Answer:
5/4.
Explanation:

Q2
$Iff\left(x\right)=x\mathrm{sin}x,thenf\text{'}\left(\frac{\mathrm{\pi}}{2}\right)isequalto$
Marks:1Answer:
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$1$
Explanation:
$Wehave,f\left(x\right)=x\mathrm{sin}x$
$\Rightarrow f\text{'}\left(x\right)=\mathrm{sin}x+x\mathrm{cos}x$
$f\text{'}\left(\frac{\mathrm{\pi}}{2}\right)=\mathrm{sin}\left(\frac{\mathrm{\pi}}{2}\right)+\frac{\mathrm{\pi}}{2}\mathrm{cos}\left(\frac{\mathrm{\pi}}{2}\right)=1$

Q3Marks:1
Answer:
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Q4Marks:1
Answer:
1 – cosx
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Q5Marks:1
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