Answer:
(a) f'(1)=-4
(b) y+4x-4=0
Step-by-step explanation:
Tangent Line of a Function
Given f(x) a real differentiable function in x=a, the slope of the tangent line of the function in x=a is given by f'(x=a). Where f' is the first derivative of f.
We are given
[tex]y=x-x^5[/tex]
The derivative is
[tex]y'=1-5x^4[/tex]
(a) The slope of the tangent line at (1,0) is
[tex]f'(1)=1-5\cdot 1^4=-4[/tex]
[tex]f'(1)=-4[/tex]
(b) The equation of the tangent line can be found with the general formula of the line:
[tex]y-y_o=m(x-x_o)[/tex]
Where m is the slope and the point (xo,yo) belongs to the line. We have m=-4, xo=1, yo=0, thus
[tex]y-0=-4(x-1)[/tex]
Or, equivalently
[tex]y+4x-4=0[/tex]
