Question

I need some help with this calculus 1 question.

Answer 1

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

$y=x-x^5$

The derivative is

$y'=1-5x^4$

(a) The slope of the tangent line at (1,0) is

$f'(1)=1-5\cdot 1^4=-4$

$f'(1)=-4$

(b) The equation of the tangent line can be found with the general formula of the line:

$y-y_o=m(x-x_o)$

Where m is the slope and the point (xo,yo) belongs to the line. We have m=-4, xo=1, yo=0, thus

$y-0=-4(x-1)$

Or, equivalently

$y+4x-4=0$