What is the equation of the tangent line of #f(x) =ln(sinx/x)# at #x=pi/4#?

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Answer 1 · 2017-04-06T22:06:29

#y = (1-4/pi)(x - pi/4) + ln(2sqrt(2)/pi)#

The corresponding y coordinate is:

#y = f(pi/4)#

#y = ln(sin(pi/4)/(pi/4))#

#y = ln((sqrt2/2)/(pi/4))#

#y = ln(2sqrt(2)/pi)#

Use the chain rule and the quotient rule to compute the derivative:

#f'(x) = f(u(g/h))#

#f(u) = ln(u)#

#f'(u) = (1/u)u'#

#u = sin(x)/x#

#f'(u) = (x/sin(x))u'#

#u' = (g'h-g(h'))/h^2#

#g= sin(x)#

#g' = cos(x)#

#h = x#

#h' = 1#

#u' = (cos(x)x-sin(x))/x^2#

#f'(u) = (x/sin(x))(cos(x)x-sin(x))/x^2#

#f'(x) = cot(x)-1/x#

The slope is the derivative evaluated at #x = pi/4#:

#f'(pi/4) = cot(pi/4)-1/(pi/4)#

#f'(pi/4) = 1-4/pi#

Using the point-slope form:

#y = (1-4/pi)(x - pi/4) + ln(2sqrt(2)/pi)#