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

A Note It may be noted here that dy · Part 5

Chapter 5: CONTINUITY AND DIFFERENTIABILITY · MATHEMATCS PART-1

is defined for all positive real numbers. Taking logarithms, we have log y = log (sin x ) sin x = sin x log (sin x ) Then dy y dx = d dx (sin x log (sin x )) = cos x log (sin x ) + sin x . (sin ) x dx = cos x log (sin x ) + cos x = ( + log (sin x )) cos x Thus dx = y (( + log (sin x )) cos x ) = ( + log (sin x )) ( sin x ) sin x cos x Example For a positive constant a find dy dx , where , and t t Solution Observe that both y and x are defined for all real t ≠ . Clearly dt = t t t t d t t +  Similarly dt = a t a t dt ≠ only if t ≠ ± .

Thus for t ≠ ± , a t t t             t t a t Example Differentiate sin x w.r.t. e cos x . Solution Let u ( x ) = sin x and v ( x ) = e cos x . We want to find / / du du dx dv dv dx .

Clearly du dx = sin x cos x and dv dx = e cos x (– sin x ) = – (sin x ) e cos x Thus du dv = 2sin 2cos x e = − Miscellaneous Exercise on Chapter Differentiate w.r.t. x the function in Exercises to . . ( x – x + ) .

sin x + cos x . ( x ) cos x . sin – ( x x ), ≤ x ≤ . , – < x < .

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