189k views
3 votes
Lety = tan(5x + 5) Find the differential dy when x = 5 and dx = 0.1

Lety = tan(5x + 5) Find the differential dy when x = 5 and dx = 0.1-example-1

1 Answer

2 votes

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

User Jakko
by
4.9k points