Question

Find the derivative of x tan x with respect to x from first

Answer 1

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

Now we find the derivative of 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'=1cos2(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'=limh→0(tan(x+h)−tanxh)

y'=limh→0⎛⎜ ⎜⎝(tanx+tanh1−tanx⋅tanh)−tanxh⎞⎟ ⎟⎠using the identity for tan (a + b) from trigonometry

=limh→0⎛⎜⎝tanx+tanh−tanx+tan2(x)tanh1−tanxtanhh⎞⎟⎠

=limh→0(tanh+tan2xtanhh⋅(1−tanxtanh))

=limh→01+tan2x1−tanxtanh⋅limh→0tanhh1

Note that limh→0(tanhh)=1 because

limh→0(tanhh)=limh→0(sinhcosh⋅h)=

limh→0(sinhh)⋅limh→0(1cosh)

=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)

= limh→01+tan2x1−tanx⋅0

=1+tan2x=sec2x

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Step-by-step explanation: