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provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.4 question 26

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Answer:

\frac{d y}{d x}=\frac{\sec ^{2}(x+y)+\sec ^{2}(x-y)}{\sec ^{2}(x-y)-\sec ^{2}(x+y)}

Hint:

Use chain rule

Given:

\tan (x+y)+\tan (x-y)=1

Solution:

\tan (x+y)+\tan (x-y)=1

Differentiate w.r.t x

\frac{d(\tan (x+y))}{d x}+\frac{d(\tan (x-y))}{d x}=\frac{d(1)}{d x}

\frac{d(\tan x+y)}{d(x+y)} \times \frac{d(x+y)}{d x}+\frac{d(\tan (x-y))}{d(x-y)} \times \frac{d(x-y)}{d x}=0

                                                                                [Using chain rule and \frac{d(\operatorname{cons} \tan t)}{d x}=0]

\sec ^{2}(x+y) \times\left(\frac{d x}{d x}+\frac{d y}{d x}\right)+\sec ^{2}(x-y)\left(\frac{d x}{d x}-\frac{d y}{d x}\right)=0

\sec ^{2}(x+y)+\sec ^{2}(x+y) \frac{d y}{d x}+\sec ^{2}(x-y)-\sec ^{2}(x-y) \frac{d y}{d x}=0

\frac{d y}{d x}\left(\sec ^{2}(x+y)-\sec ^{2}(x-y)\right)=-\left(\sec ^{2}(x+y)+\sec ^{2}(x-y)\right)

\frac{d y}{d x}=\frac{\sec ^{2}(x+y)+\sec ^{2}(x-y)}{\sec ^{2}(x-y)-\sec ^{2}(x+y)}

 

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