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Differentiate w.r.t. x the function

Differentiate w.r.t. x the function
Using chain rule, we obtain

Differentiate w.r.t. x the function
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Taking logarithm on both the sides, we obtain
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Differentiating both sides with respect to x, we obtain

Differentiate w.r.t. x the function
![]()
Using chain rule, we obtain

Differentiate w.r.t. x the function

Differentiate w.r.t. x the function
![]()
Taking logarithm on both the sides, we obtain
![]()
Differentiating both sides with respect to x, we obtain

Differentiate w.r.t. x the function
, for some constant a and b.
![]()
By using chain rule, we obtain

Differentiate w.r.t. x the function
![]()
Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

Differentiate w.r.t. x the function
, for some fixed
and


Differentiating both sides with respect to x, we obtain


Differentiating both sides with respect to x, we obtain

s = aa
Since a is constant, aa is also a constant.
∴![]()
From (1), (2), (3), (4), and (5), we obtain

Differentiate w.r.t. x the function
, for

Differentiating both sides with respect to x, we obtain

Differentiating with respect to x, we obtain

Also,

Differentiating both sides with respect to x, we obtain

Substituting the expressions of
in equation (1), we obtain

Find
, if

Find
, if

If
, for, −1 < x <1, prove that
It is given that,
![]()

Differentiating both sides with respect to x, we obtain

Hence, proved.
If
, for some
prove that
is a constant independent of a and b.
It is given that,![]()
Differentiating both sides with respect to x, we obtain


Hence, proved.
If
with
prove that

Then, equation (1) reduces to.png)
⇒sin(a+y−y)
Hence, proved.
If
and
, find


If
, show that
exists for all real x, and find it.
It is known that, ![]()
Therefore, when x ≥ 0, ![]()
In this case,
and hence, ![]()
When x < 0, ![]()
In this case,
and hence, ![]()
Thus, for
,
exists for all real x and is given by,
![]()
Using mathematical induction prove that
for all positive integers n.
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For n = 1,
![]()
∴P(n) is true for n = 1
Let P(k) is true for some positive integer k.
That is, ![]()
It has to be proved that P(k + 1) is also true.

Thus, P(k + 1) is true whenever P (k) is true.
Therefore, by the principle of mathematical induction, the statement P(n) is true for every positive integer n.
Hence, proved.
Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
![]()
Differentiating both sides with respect to x, we obtain

Does there exist a function which is continuos everywhere but not differentiable at exactly two points? Justify your answer ?

.png)
It can be seen from the above graph that, the given function is continuos everywhere but not differentiable at exactly two points which are 0 and 1.
If
, prove that

Thus, 
If
, show that ![]()
It is given that,![]()


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