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23 January, 06:05

Derivative of tan (2x+3) using first principle

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  1. 23 January, 06:08
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    There are two possible answers depending upon what you mean by "first principles."

    1) If by first principles, you mean using derivatives we have already found (like the derivative of y = sin x and y = cos x) and rules we already know (like the quotient rule) this is relatively straightforward.

    We know from trigonometry that y = tan (x) = sin (x) cos (x)

    Now we find the derivative of this expression using the quotient rule:

    y' = (cos (x) ⋅ cos (x) - sin (x) ⋅ ( - sin (x)) cos2 (x))

    y' = cos2 (x) + sin2 (x) cos2 (x)

    y' = 1 cos2 (x) = sec2 (x)

    2) If by first principles, you mean that you would like to go back to the definition of the derivative, then we have to do a bit more work.

    y' = lim h→0 (tan (x+h) - tanx h)

    y' = lim h→0 ⎛⎜⎝ (tanx + tanh 1 - tanx ⋅ tanh) - tanx h ⎞⎟⎠ using the identity for tan (a + b) from trigonometry

    = lim h→0 ⎛⎜⎝ tanx + tanh - tanx + tan2 (x) tanh 1 - tanx tanh h ⎞⎟⎠

    = lim h→0 (tanh + tan2 x tanh h⋅ (1 - tanx tanh))

    = lim h→0 1 + tan2 x 1 - tanx tanh ⋅ lim h→0 tanh h 1

    Note that lim h→0 (tanh h) = 1 because

    lim h→0 (tanh h) = lim h→0 (sinh cosh ⋅h) =

    lim h→0 (sinh h) ⋅ lim h→0 (1 cosh)

    =1*1=1 (the first limit is a famous one, proven by the squeeze theorem) that you probably learned in your calculus class, while the second limit can be found by simply substituting zero for h)

    = lim h→0 1 + tan2 x 1 - tanx ⋅0

    =1 + tan2 x = seconds
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