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Learn Miscellaneous Exercise 5 with Free Lessons & Tips

Differentiate w.r.t. x the function 

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If, show that 

It is given that,



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Differentiate w.r.t. x the function 

Using chain rule, we obtain

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Differentiate w.r.t. x the function

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

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Differentiate w.r.t. x the function 

Using chain rule, we obtain

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Differentiate w.r.t. x the function

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Therefore, equation (1) becomes

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Differentiate w.r.t. x the function 

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

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 Differentiate w.r.t. x the function for some constant a and b.

By using chain rule, we obtain

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Differentiate w.r.t. x the function 

Taking logarithm on both the sides, we obtain

Differentiating both sides with respect to x, we obtain

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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

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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

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Find, if 

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Find, if 

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If, for, −1 < x <1, prove that

It is given that,

Differentiating both sides with respect to x, we obtain

Hence, proved.

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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.

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If  with prove that

Then, equation (1) reduces to

sin(a+yy)  


Hence, proved.

 

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If and, find 

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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,

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Using mathematical induction prove that for all positive integers n.

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.

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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

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


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.

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If, prove that 

Thus,

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