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

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

Answers (1)

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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