Differentiate and find the equation of the tangent and the normal. `y=sqrt(2x+3) `  at the point x=3



1 Answer | Add Yours

mjripalda's profile pic

Posted on (Answer #1)

First, plug-in x=3 to the equation of the curve to get the corresponding value of y.


So, the point of tangency is (3, 3).

Next, determine the slopes of the tangent and the normal line.

To do so, take the derivative of the given curve.



To get the derivative, express the radical as exponent and apply the power rule which is `(u^n)'=n*u^(n-1)*u'` .


`y'=1/2(2x+3)^(-1/2) *(2x+3)'`



To simplify, apply the negative exponent rule which is `a^(-m)=1/a^m` .




Note that the slope of a tangent at a certain point in the curve is equal to y' `( m_T=y')` .

So, plug-in x=3 to y'.


Hence, the slope of the tangent line at point (3,3) is `m_T=1/3` .

Since a normal line is perpendicular to tangent line, its slope is:


Now that the point of tangency and the slopes of the two lines are known, use the point-slope formula to get the equation of each line.

The formula is:


For the tangent line, plug-in `m_T=1/3` and the point (3,3) to the formula.






Hence, the equation of the tangent line is `y=x/3+2` .

For the normal line, substitute `m_N=-3` and the point (3,3) to the same formula.





Thus, the equation of the normal line is `y=-3x+12` .

We’ve answered 395,995 questions. We can answer yours, too.

Ask a question