Please help?

Question

Consider the function f(x)=(4x^2-3x+4)^cos(x) Use logarithmic differentiation to find f'(1).....

Answer 1

Take log of both sides:

ln[f(x)] = cos(x)*ln(4x^2-3x+4)

differentiate both sides (use product rule and chain rule):

1/f(x) * f'(x) = -sin(x)*ln(4x^2-3x+4) + cosx * 1/([4x^2-3x+4) * (8x-3)

solve for f'(x) [= df(x)/dx]:

f'(x) = f(x)*{-sin(x)*ln(4x^2-3x+4) + cosx * 1/([4x^2-3x+4) *(8x-3)}

put x = 1 into the above, noting that f(1) = 5^cos(1):

5^cos(1) * [-sin(1) * ln(5) + cos(1)*(1/5)*5]

5*cos(1) * [cos(1) - ln(5)*sin(1)

sin(1) = 0.8414; cos(1) = 0.5403; ln(5) = 1.609

Answer 2

5.232 to the second power