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Solve the given problems by using implicit differentiation.At what point(s) is the tangent to the curve $y^{2}=2 x^{3}$ perpendicular to the line $4 x-3 y+1=0 ?$

00:55

Frank L.

Calculus 1 / AB

Chapter 23

The Derivative

Section 8

Differentiation of Implicit Functions

Derivatives

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we're going to find the equation of the tangent line to the curve y equals square root one plus X cubed at the 10.23 The slope of the tangent line will be the derivative at that point. So let's start by finding the derivative. And I'm going to rewrite why, as one plus X cubed to the 1/2 power. And so then I can use the chain rule, starting with the derivative of the 1/2 power function. Bring down the 1/2 and raise one plus X cubed to the negative 1/2 power. And then we multiply that by the derivative of the inside, the derivative of one plus x cubed is three x squared. Now we're evaluating this derivative at X equals two. We could simplify first, or we could substitute the two in first and simplify after. So that's what I'm going to do. So 1/2 times one plus two cubed to the negative 1/2 times three times two squared. Okay, so this gives us 1/2 times nine to the negative 1/2 times 12 and nine to the negative. 1/2 is 1/3 so we have 1/2 times 1/3 times 12 So that is just too. So are slope is, too. Now we can use point slope form. Why minus y one equals m times X minus X one. To find the equation of the line where the point x one y one is 23 So we have why minus three equals air slope to times the quantity X minus two and we can simplify that and we'll have the equation of the line. So let's distribute the two. So why minus three equals two X minus four and then at three to both sides and we have y equals two x minus one.

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