Given:
[tex]y=tan\left(2x+4\right)[/tex]
To find:
The differential when x = 4 and dx = 0.1.
Explanation:
Differentiating with respect to x we get,
[tex]\begin{gathered} \frac{dy}{dx}=sec^2\left(2x+4\right)\cdot(2) \\ \frac{dy}{dx}=2sec^2\left(2x+4\right) \\ dy=2sec^2\left(2x+4\right)dx.............(1) \end{gathered}[/tex]
Substituting x = 4 and dx = 0.1 in equation (1), we get
[tex]\begin{gathered} dy=2sec^2(2(4)+4)\cdot(0.1) \\ dy=0.2sec^2(12) \end{gathered}[/tex]
Therefore, the differential is,
[tex]dy=0.2sec^{2}(12)[/tex]
Final answer:
The differential is,
[tex]dy=0.2sec^{2}(12)[/tex]
