How do you find the equation of the tangent line to the curve #y=1/(sinx+cosx)# at #(0,1)#?

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Answer 1 · 2017-08-29T21:45:09

The equation of the tangent is #y = 1 - x#

This can be rewritten as

#y = (sinx + cosx)^-1#

Letting #u = sinx + cosx#, we have:

#y' = cosx - sinx * -1/u^2 = -(cosx - sinx)/(sinx + cosx)^2#

#y' = (sinx -cosx)/(sinx + cosx)^2#

At #(0, 1)#, the tangent will have slope

#y'(0) = (sin(0) - cos(0))/(sin(0) + cos(0)^2)#

#y'(0) = (0 - 1)/(0 + 1)^2#

#y'(0) = -1/1#

#y'(0) = -1#

Now by point-slope form, we can deduce the equation of the tangent line.

#y - y_1 = m(x- x_1)#

#y - 1 = -1(x - 0)#

#y = -x + 1#

Hopefully this helps!