Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):

$$\dfrac{cosx}{1+sinx}$$

Asked by Abhisek | 1 year ago |  51

Solution :-

Let f(x) = $$\dfrac{cosx}{1+sinx}$$

By quotient rule,

$$=\dfrac{(1+sinx)\dfrac{d}{dx}(cosx)-(cosx)\dfrac{d}{dx}(1+sinx)}{(1+sinx)^2}$$

$$\dfrac{(1+sinx)(-sinx)-(cosx)(cosx)}{(1+sinx)^2}$$

$$\dfrac{-sinx-sin^2x-(cos^2x)}{(1+sinx)^2}$$

$$\dfrac{-sinx-(sin^2x+cos^2x)}{(1+sinx)^2}$$

$$\dfrac{-(1-sinx)}{(1+sinx)^2}$$

$$\dfrac{-1}{ (1+sinx)^2}$$

Answered by Pragya Singh | 1 year ago

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