Thread: To find the differential dy, solved alread, need to chek with your expertize :)

I have the function y = tan(3x + 6); need to find dy when x=3 and dx=0.03

By the formula dy = f'(x)*dx ;

f'(x)= sec^2(3x+6)(4)dx = (1/cos^2(3x+6))(4)dx, (need to calculate in radians)

so,forf'(3) = (1/cos^2(3x+6)) = (1/cos^2(3(3)+6)) = (1/cos^2(15)) = (1/cos^2(45-30)).... am I on the right track , i'm confused on which identity to apply further

Just give me a hint, I don't need the solution,
Thank you

10 viewed and no reply - does this mean this is correct ? thank you

Originally Posted by dokrbb

I have the function y = tan(3x + 6); need to find dy when x=3 and dx=0.03

By the formula dy = f'(x)*dx ;

f'(x)= sec^2(3x+6)(3)dx = (1/cos^2(3x+6))(3)dx, (need to calculate in radians)

so,for f'(3) = (1/cos^2(3x+6)) = (1/cos^2(3(3)+6)) = (1/cos^2(15)) = (1/cos^2(45-30)).... am I on the right track , i'm confused on which identity to apply further

Thank you

I realized I did completely wrong, I corrected all the mistakes but I'm confused on which identity to apply further

Just give me a hint, I don't need the solution,

I continued in this way:

f'(3) = (1/cos^2(3x+6)) = (1/cos^2(3(3)+6)) = (1/cos^2(15)) = (1/[(1+cos2(15)/2)]) = (2/(1+cos30)) = {2/[1+ sqrt(3)/2]} = [2/[(2+sqrt(3))/2]] = 4/((2+sqrt(3)),

so, dy = (4/((2+sqrt(3)))(3)(0.03) = 0.36/(2+sqrt(3)) = ... am I correct this time, please, tell me I am