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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 22: Quotient rule

# Tangent to y=𝑒ˣ/(2+x³)

Sal finds the equation of the line tangent to the curve y=eˣ/(2+x³) at the point (1,e/3). Created by Sal Khan.

## Want to join the conversation?

• This video refers to the chain rule, but if you follow 'Calculus 1', the chain rule will only be explained in the next section.
• That is fine.
You can do this perfectly using just the quotient rule.
• Could you have used the quotient rule here? Curious because it wasn't used but is in the quotient rule section.
Would have appreciated more direct uses of the quotient rule here (in these videos) instead of going by way of the product and chain rules.
• i used the quotient rule and got the same asnwer
• when tangent line is parallel to y-axis its slope will be?
• The slope of vertical line (a line parallel to the y axis) is undefined. Why? Because slope is defined as change in y divided by change in x. If the line is vertical there is no change in x, that is given that change in x is x(1)-x(0), then x(1)-x(0) = 0 because x(1) = x(2). In that case the calculation for slope results in division by zero, which is undefined.
• starting at around it seems sal forgot to multiply in x^2? im not sure if i missed where he explained it, but i beleive it should have been e/3 + (e(x^2)/3). i appologize if i am mistaken
• No, if you put in x=1 you get x^2=1.
So he just multiplied by 1.
• sir
as we say that the slope of a normal line at a point is the negative reciprocal of tangent line how we can interprate that , does it mean that the slope of normal line is equal to - deltax /delta y ??
• The normal line is perpendicular to the tangent line. So go back to algebra. Remember two parallel lines have the same slope. Lines that are perpendicular have slopes that are negative reciprocals of each other. So the slope of the normal line is the negative reciprocal of the slope of the tangent line.

If you are determining -deltax/deltay by finding the derivative of the function (deltay/deltax) and then taking the negative reciprocal, you are fine. But if you try to find -deltax/deltay by taking the negative derivative of the inverse function, that doesn't work.
• why is the derivative of (e power x ) also e power x ?
• What if we want to find the equation of the normal line to the given curve at x = 1 ? as the slope of the tangent line equals 0 at x =1 , the slope of the normal line which is the negative reciprocal of the slope of the tangent line becomes negative infinity. so the usual slope intercept form of the equation of a line does not work for the normal line here. but since the tangent line at x = 1 is parallel to the x-axis, can we say that the normal line at x = 1 will be parallel to the y axis ? and since it passes through the point (1,e/3) , can the equation of the normal line be written as x = 1 ?
• That is correct. You cannot legitimately calculate using the slope of a vertical line, since the slope is infinite. However, you can observe that the normal line is vertical and write the appropriate equation for a vertical line: x=a, or in this case, x=1.
• So seeing as this is the Quotient rule section I tried to solve using the quotient rule, here's what I got :
d/dx e^x/(2+x^3) = d/dx(e^x) (2+x^3) - d/dx(2+x^3) (e^x) / (2+x^3)^2
= (e^x) (2+x^3) - (3x^2) (e^x) / (2+x^3)^2
substituting x=1
= e(2+1) - 3(e) / (2+1)^2
= 3e -3e / 9
= 0 / 9
= 0