Question

Lety = tan(3x + 3) . Find the differential dy when x = 5 and dx = 0.4

Answer 1

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.

`\begin{gathered} y=tan(3x+3) \\ dy=3sec^2(3x+3)dx \end{gathered}`

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.

`dy=\lbrace3sec^2[(3(5)+3)]\rbrace(0.4)`

Then, simplify.

`dy=[3sec^2(18)](0.4)`
`dy=3.3167(0.4)`
`dy=1.32668\approx1.33`

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.

`dy=[3sec^2(18)](0.8)`
`dy=3.3167(0.8)`
`dy=2.6534\approx2.65`

Hence, at x = 5 and dx = 0.8, the differential dy is approximately equal to 2.65.