Determine an equation of the line tangent to the graph

dmcalvin

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Determine an equation of the line tangent to the graph y=x4 divided by (x2-6)5 at the point (3,1/3)Where F(x)=x4 and G(x)=(x2-2)5

The point-slope-formula for a line is given by y-y1=m(x-x1). To use this formula, first determine the slope of the tangent line at a given point. The slope of the line tangent to the graph of a function is the derivative of the function. Use the quotient and chain rule to find the derivative of the equation.

f '(x)= (d/dx)(x4)= 4x3
g '(x)= (d/dx)(x2-2)5 here one needs to use the chain rule to find the derivative of g(x)
in this instance g(x)=x2-2 and f(u)=u5
f '(u)=(d/du)(u5)= 5u4
g '(x)=(d/dx)(x2-2)= 2x
together they equal (5u4)(2x)
= 5(x2-2)4(2x) = g '(x) which is the derivative needed for the quotient formula
the quotient formula is (g(x))(f '(x))-(g '(x))(f(x)) all divided by ((g(x))2
put in the two bold derivatives where f '(x)= 4x3 and g '(x)=5(x2-2)4(2x)


you get (x2-2)5*(4x3)-x5*5(x2-2)4(2x) divided by ((x2-2)5)2
I think

the simplified derivative = ?
Can somebody show me step-by-step how to simplify this?
it should look something like (?-8x3) divided by (x2-2)6
Also can somebody explain why the "-x5" was placed into the quotient formula?
 

Determine an equation of the line tangent to the graph y=x4 divided by (x2-6)5 at the point (3,1/3)Where F(x)=x4 and G(x)=(x2-2)5

The point-slope-formula for a line is given by y-y1=m(x-x1). To use this formula, first determine the slope of the tangent line at a given point. The slope of the line tangent to the graph of a function is the derivative of the function. Use the quotient and chain rule to find the derivative of the equation.

f '(x)= (d/dx)(x4)= 4x3
g '(x)= (d/dx)(x2-2)5 here one needs to use the chain rule to find the derivative of g(x)
in this instance g(x)=x2-2 and f(u)=u5
f '(u)=(d/du)(u5)= 5u4
g '(x)=(d/dx)(x2-2)= 2x
together they equal (5u4)(2x)
= 5(x2-2)4(2x) = g '(x) which is the derivative needed for the quotient formula
the quotient formula is (g(x))(f '(x))-(g '(x))(f(x)) all divided by ((g(x))2
put in the two bold derivatives where f '(x)= 4x3 and g '(x)=5(x2-2)4(2x)


you get (x2-2)5*(4x3)-x5*5(x2-2)4(2x) divided by ((x2-2)5)2
I think

the simplified derivative = ?
Can somebody show me step-by-step how to simplify this?
it should look something like (?-8x3) divided by (x2-2)6
Also can somebody explain why the "-x5" was placed into the quotient formula?
To begin with, I would avoid the quotient rule, in favor of multiplication by a negative power. BTW, why did you change (x^2-6) to (x^2-2)? To get the point (3, 1/3) the function must be
\(\displaystyle \displaystyle y = x^4\ (x^2 - 6)^{-5}\)

\(\displaystyle \displaystyle \dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x^3(x^2 - 6)^{-5} - 5x^4\ (x^2 - 6)^{-6}\ (2x) \)

.......\(\displaystyle \displaystyle= \dfrac{4x^3(x^2 - 6) - 10 x^5}{(x^2 - 6)^6}\)

.......\(\displaystyle \displaystyle= \dfrac{-6 x^5 - 24 x^3}{(x^2 - 6)^6}\)

The x^5 that you question looks like it should be x^4, the original numerator. To simplify the expression you got, begin by cancelling (x^2 - 6)^4 from numerator and denominator. Then combine like powers of x in the numerator. We should wind up with the same result.
 
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To begin with, I would avoid the quotient rule, in favor of multiplication by a negative power. BTW, why did you change (x^2-6) to (x^2-2)? To get the point (3, 1/3) the function must be
\(\displaystyle \displaystyle y = x^4\ (x^2 - 6)^{-5}\)

\(\displaystyle \displaystyle \dfrac{\mathrm{d}y}{\mathrm{d}x} = 4x^3(x^2 - 6)^{-5} - 5x^4\ (x^2 - 6)^{-6}\ (2x) \)

.......\(\displaystyle \displaystyle= \dfrac{4x^3(x^2 - 6) - 10 x^5}{(x^2 - 6)^6}\)

.......\(\displaystyle \displaystyle= \dfrac{-6 x^5 - 24 x^3}{(x^2 - 6)^6}\)

The x^5 that you question looks like it should be x^4, the original numerator. To simplify the expression you got, begin by cancelling (x^2 - 6)^4 from numerator and denominator. Then combine like powers of x in the numerator. We should wind up with the same result.

I am sorry about the change but I was looking at several of them at once and I must have put in the wrong one. I believe with the numbers I used the problem was (x^4)/(x^2-2)^5 with the points (2,1/2) Also I am currently learning to solve derivatives with the quotient rule and part of the answer requires that I use it. However, your way presumably gives the right answer so I could figure out how to solve the problem using your method. Unfortunately with the problem being (x^4)/(x^2-2)^5 your answer wouldn't be correct since you used (x^2-6). Again, I am sorry about that.

EDIT* Thank you for your patience and your help with this particular problem. Using you method of multiplication of a negative power I was able to correctly solve the problem with an answer of
(-6x^5-8x^3)/(x^2-2)^6
 
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I am sorry about the change but I was looking at several of them at once and I must have put in the wrong one. I believe with the numbers I used the problem was (x^4)/(x^2-2)^5 with the points (2,1/2) Also I am currently learning to solve derivatives with the quotient rule and part of the answer requires that I use it. However, your way presumably gives the right answer so I could figure out how to solve the problem using your method. Unfortunately with the problem being (x^4)/(x^2-2)^5 your answer wouldn't be correct since you used (x^2-6). Again, I am sorry about that.

EDIT* Thank you for your patience and your help with this particular problem. Using you method of multiplication of a negative power I was able to correctly solve the problem with an answer of
(-6x^5-8x^3)/(x^2-2)^6
YES :grin: I had that answer written down, but couldn't make it work at point (3, 1/3).

To find slope m, evaluate the derivative at x=2. I get m=-4.
Then make the line go through the point (2, 1/2).
 
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