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?