Question

Lety = tan(5x + 5) Find the differential dy when x = 5 and dx = 0.1

Answer 1

Given:

The expression is:

`y=\tan(5x+5)`

Find-:

(a)

Differential dy when x = 5 and dx = 0.1

(b)

Differential dy when x = 5 and dx = 0.1

Explanation-:

Differential is,

`\begin{gathered} y=\tan(5x+5) \\ \\ \end{gathered}`

The formula of the differential is:

`(d)/(d\theta)\tan\theta=\sec^2\theta`

Apply the formula then,

`\begin{gathered} y=\tan(5x+5) \\ \\ (dy)/(dx)=(d)/(dx)\tan(5x+5) \\ \\ (dy)/(dx)=\sec^2(5x+5)\cdot(d)/(dx)(5x+5) \\ \\ (dy)/(dx)=5\sec^2(5x+5) \end{gathered}`

The given value is:

`\begin{gathered} x=5 \\ \\ dx=0.1 \end{gathered}`

So, the value of dy is:

`\begin{gathered} (dy)/(dx)=5\sec^2(5x+5) \\ \\ dy=5\sec^2(5*5+5)* dx \\ \\ dy=5\sec^230*0.1 \\ \\ dy=0.5\sec^230 \end{gathered}`

Value is:

`\begin{gathered} dy=0.5*((2)/(√(3)))^2 \\ \\ dy=0.5*(4)/(3) \\ \\ dy=(2)/(3) \\ \\ dy=0.667 \end{gathered}`

Value of dy is 0.667

(b)

The value of dy is:

`dy=5\sec^2(5x+5)* dx`
`\begin{gathered} x=5 \\ \\ dx=0.2 \end{gathered}`

Value of dy is:

`\begin{gathered} dy=5\sec^2(5*5+5)*0.2 \\ \\ dy=1*\sec^230 \\ \\ dy=((2)/(√(3)))^2 \\ \\ dy=(4)/(3) \\ \\ dy=1.333 \end{gathered}`

Value of dy is 1.333