Applying the derivative of the trigonometric functions, the derivative of tan x is sec²x.
Hence, the derivative of tan (3x + 3 ) is sec² (3x + 3) times the derivative of 3x - 3 which is 3.
[tex]\begin{gathered} y=tan(3x+3) \\ dy=3sec^2(3x+3)dx \end{gathered}[/tex]
To find the differential dy when x = 5 and dx = 0.4, simply replace the x and dx in the dy function with their given values.
[tex]dy=\lbrace3sec^2[(3(5)+3)]\rbrace(0.4)[/tex]
Then, simplify.
[tex]dy=[3sec^2(18)](0.4)[/tex][tex]dy=3.3167(0.4)[/tex][tex]dy=1.32668\approx1.33[/tex]
Hence, at x = 5 and dx = 0.4, the differential dy is approximately equal to 1.33.
For x = 5 and dx = 0.8, we do the same process above but this time, multiply the derivative by 0.8.
[tex]dy=[3sec^2(18)](0.8)[/tex][tex]dy=3.3167(0.8)[/tex][tex]dy=2.6534\approx2.65[/tex]
Hence, at x = 5 and dx = 0.8, the differential dy is approximately equal to 2.65.
