Given:
The expression is:
[tex]y=\tan(5x+5)[/tex]
Find-:
(a)
Differential dy when x = 5 and dx = 0.1
(b)
Differential dy when x = 5 and dx = 0.1
Explanation-:
Differential is,
[tex]\begin{gathered} y=\tan(5x+5) \\ \\ \end{gathered}[/tex]
The formula of the differential is:
[tex]\frac{d}{d\theta}\tan\theta=\sec^2\theta[/tex]
Apply the formula then,
[tex]\begin{gathered} y=\tan(5x+5) \\ \\ \frac{dy}{dx}=\frac{d}{dx}\tan(5x+5) \\ \\ \frac{dy}{dx}=\sec^2(5x+5)\cdot\frac{d}{dx}(5x+5) \\ \\ \frac{dy}{dx}=5\sec^2(5x+5) \end{gathered}[/tex]
The given value is:
[tex]\begin{gathered} x=5 \\ \\ dx=0.1 \end{gathered}[/tex]
So, the value of dy is:
[tex]\begin{gathered} \frac{dy}{dx}=5\sec^2(5x+5) \\ \\ dy=5\sec^2(5\times5+5)\times dx \\ \\ dy=5\sec^230\times0.1 \\ \\ dy=0.5\sec^230 \end{gathered}[/tex]
Value is:
[tex]\begin{gathered} dy=0.5\times(\frac{2}{\sqrt{3}})^2 \\ \\ dy=0.5\times\frac{4}{3} \\ \\ dy=\frac{2}{3} \\ \\ dy=0.667 \end{gathered}[/tex]
Value of dy is 0.667
(b)
The value of dy is:
[tex]dy=5\sec^2(5x+5)\times dx[/tex][tex]\begin{gathered} x=5 \\ \\ dx=0.2 \end{gathered}[/tex]
Value of dy is:
[tex]\begin{gathered} dy=5\sec^2(5\times5+5)\times0.2 \\ \\ dy=1\times\sec^230 \\ \\ dy=(\frac{2}{\sqrt{3}})^2 \\ \\ dy=\frac{4}{3} \\ \\ dy=1.333 \end{gathered}[/tex]
Value of dy is 1.333
