Find the derivative of the functions sec x

Asked by Abhisek | 1 year ago |  79

##### Solution :-

Let f(x) =  sec x

$$\dfrac{1}{cosx}$$

By differentiating both sides, we get

$$f'(x) =\dfrac{d}{dx}(\dfrac{1}{cosx})$$

$$f'(x) =cosx\dfrac{cosx\dfrac{d}{dx}(1)-1\dfrac{d}{dx}(cosx)}{cos^2x}$$

$$\dfrac{cosx\times 0-(-sinx)}{cos^2x}$$

We get

$$\dfrac{sinx}{cos^2x}= \dfrac{sinx}{cosx}\times \dfrac{1}{cosx}$$

= tan x sec x

Answered by Pragya Singh | 1 year ago

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