📖 generic · CBSE Class 12th English Medium · MATHEMATCS PART-1 · Page 33question

A Note It may be noted here that dy · Part 4

Chapter 5: CONTINUITY AND DIFFERENTIABILITY · MATHEMATCS PART-1

(i) cos – (sin x ) (ii) tan −  (iii) −  Solution (i) Let f ( x ) = cos – (sin x ). Observe that this function is defined for all real numbers. We may rewrite this function as f ( x ) = cos – (sin x ) = cos         π = π − Thus f ′ ( x ) = – . (ii) Let f ( x ) = tan – .

Observe that this function is defined for all real numbers, where cos x ≠ – ; i.e., at all odd multiplies of π . We may rewrite this function as f ( x ) = tan −  sin tan 2cos tan tan −  = Observe that we could cancel cos  in both numerator and denominator as it is not equal to zero. Thus f ′ ( x ) = . (iii) Let f ( x ) = sin – .

To find the domain of this function we need to find all x such that −≤ . Since the quantity in the middle is always positive, we need to find all x such that , i.e., all x such that x + ≤ + x . We may rewrite this as ≤ x + x which is true for all x . Hence the function is defined at every real number.

By putting x = tan θ , this function may be rewritten as f ( x ) = −  = sin − + ( 2tan tan = sin – [sin θ ] = θ = tan – ( x ) Thus f ′ ( x ) = ( ) ( )log2 x ⋅ log2 Example Find f ′ ( x ) if f ( x ) = (sin x ) sin x for all < x < π . Solution The function y = (sin x ) sin x

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