# Let y=tan(5x+2). Find the differential dy when x=2 and dx=0.4Find the differential dy when x=2 and dx=0.8

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You need to remember that the tangent function is rational, hence you should write the tangent function as `tan(5x+2) = sin(5x+2)/cos(5x+2)` .

You need to find dy, hence you need to differentiate with respect to x using quotient rule such that:

`dy = (sin(5x+2)/cos(5x+2))dx`

`dy = (5cos(5x+2)*cos(5x+2)+ 5 sin(x+2)*sin(x+2))/(cos^2(5x+2))dx`

You need to factor out 5 such that:

`dy = 5(cos^2(5x+2) + sin^2(x+2))/(cos^2(5x+2))dx`

You need to use the basic formula of trigonometry such that:

`cos^2(5x+2) + sin^2(x+2) = 1`

`dy = 5/(cos^2(5x+2))dx`

You need to substitute 2 for x and `0.4` for `dx` in `dy = 5/(cos^2(5x+2))dx` such that:

`dy = 5/(cos^2(5*2+2))*0.4`

`dy = 5*4/(10(cos^2(12))) =gt dy = 2/(cos^2(12))`

You need to substitute 2 for x and 0.8 for`dx` in `dy = 5/(cos^2(5x+2))dx` such that:

`dy = 5/(cos^2(5*2+2))*0.8`

`dy = 5*8/(10(cos^2(12))) =gt dy = 4/(cos^2(12))`

**Hence, evaluating `dy` at `x=2` and `dx=0.4;dx=0.8` yields `dy = 2/(cos^2(12))|_(x=2,dx=0.4)` and `dy = 4/(cos^2(12))|_(x=2,dx=0.8).` **

The function `y = tan(5x + 2)`

`dy` = `5*sec^2(5x + 2) dx`

For x = 2 and `dx` = 0.4, `dy` = `5*sec^2(12)*0.4`

=> ` 2*sec^2 12`

For x = 2 and `dx` = 0.8, `dy` = `5*sec^2(12)*0.8`

=> `4*sec^2 12`

**The value of `dy` when x = 2 and `dx` = 0.4 is `2*sec^2 12` and when x = 2 and `dx` = 0.8 it is **`4*sec^2 12`