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Continuity And Differentiability

Question

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Differentiate the following functions w.r.t.x:
$tan left parenthesis straight x plus straight y right parenthesis plus tan left parenthesis straight x minus straight y right parenthesis equals 1$

Solution

Here space space tan left parenthesis straight x plus straight y right parenthesis plus tan left parenthesis straight x minus straight y right parenthesis equals 1 Differentiating space straight w. straight r. straight t. straight x comma space sec squared left parenthesis straight x plus straight y right parenthesis open square brackets 1 plus dy over dx close square brackets plus sec squared left parenthesis straight x minus straight y right parenthesis open square brackets 1 minus dy over dx close square brackets equals 0 therefore space sec squared left parenthesis straight x plus straight y right parenthesis plus sec squared left parenthesis straight x plus straight y right parenthesis dy over dx plus sec squared left parenthesis straight x minus straight y right parenthesis minus sec squared left parenthesis straight x minus straight y right parenthesis dy over dx equals 0 therefore space left square bracket sec squared left parenthesis straight x plus straight y right parenthesis plus sec squared left parenthesis straight x plus straight y right parenthesis right square bracket dy over dx equals negative left square bracket sec squared left parenthesis straight x plus straight y right parenthesis minus sec squared left parenthesis straight x plus straight y right parenthesis right square bracket therefore space dy over dx equals negative open square brackets fraction numerator sec squared left parenthesis straight x plus straight y right parenthesis plus sec squared left parenthesis straight x plus straight y right parenthesis over denominator sec squared left parenthesis straight x plus straight y right parenthesis minus sec squared left parenthesis straight x plus straight y right parenthesis end fraction close square brackets

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