[tex]\large\underline{\mathbb{QUESTION}:}[/tex]
Find the slope of the tangent line to the graph of x=–7y³ + 6y – 3 at (–2,–1).
[tex]\large\underline{\mathbb{ANSWER}:}[/tex]
[tex]\large\underline{\mathbb{SOLUTION}:}[/tex]
A function y of a variable x may be given in one of two ways: explicitly or implicitly.
An explicit function y of a variable x is defined by an equation of the form y = f(x). An implicit function y of a variable x satisfies an equation between x and y in which y is not isolated.
The derivative of a function provides a rule to find the slope of the tangent line to the graph of that function.
To differentiate an implicit function y with respect to x:
To differentiate yⁿ, use this formula:
The slope of the tangent line to a graph at a point is the value of y′ at that point.
The equation x = –7y³ + 6y – 3 defines y as an implicit function of x since y is not isolated. To find y′, first differentiate both sides of the equation.
[tex]\sf \: x=7y^{3} +6y-3[/tex]
[tex]\sf \: (x)' =(7y^{3} +6y-3)'[/tex]
[tex]\sf \: (x)'=(7y^{3})' +(6y)'-(3)'[/tex]
[tex]\sf \: 1=-21y^{2} y'+6y'[/tex]
Next, plug in x = –2 and y = –1. Then, solve for y′.
[tex]\sf \: 1=-21(1)^{2} y'+6y'[/tex]
[tex]\sf \: 1=-15y'[/tex]
[tex]\sf \: -\dfrac{1}{15} =y'[/tex]
So the slope of the tangent line is [tex]\sf \: -\dfrac{1}{15}[/tex].
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