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Q.Find the derivative of sec⁡3x\sec 3x from the first principle.

Andhra Pradesh BieapBIEAP Intermediate Board (1st Year) 2023Subjective· 4mImportance★★★★★
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Form the first-principles difference quotient, convert the cosine difference to a product using the sum-to-product identity, then take the limit.

Let f(x)=sec⁡3x=1cos⁡3xf(x)=\sec3x=\dfrac{1}{\cos3x}.

f(x+h)−f(x)=1cos⁡3(x+h)−1cos⁡3x=cos⁡3x−cos⁡3(x+h)cos⁡3(x+h)cos⁡3xf(x+h)-f(x) = \dfrac{1}{\cos3(x+h)}-\dfrac{1}{\cos3x} = \dfrac{\cos3x-\cos3(x+h)}{\cos3(x+h)\cos3x}

Using cos⁡C−cos⁡D=2sin⁡C+D2sin⁡D−C2\cos C-\cos D = 2\sin\dfrac{C+D}{2}\sin\dfrac{D-C}{2} with C=3x, D=3x+3hC=3x,\ D=3x+3h:

cos⁡3x−cos⁡3(x+h)=2sin⁡ ⁣(3x+3h2)sin⁡ ⁣(3h2)\cos3x-\cos3(x+h) = 2\sin\!\left(3x+\dfrac{3h}{2}\right)\sin\!\left(\dfrac{3h}{2}\right)

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