Find Equation of Tangent Line

[nycmathguy]

Calculus
Section 3.1

Can you do 40 as a guide for me to do the rest?
Show steps along the way.

Thank you.

 

[MathLover1]

40.

y=x^(1/4) -x at point (1,0)

first find the slope of tangent line

(dy/dx)(x^(1/4) -x )= 1/(4x^(3/4)) - 1

compute the slope at point (1,0)

m=1/(4*1^(3/4)) - 1= -3/4

so far, tangent line is: y= -(3/4)x +b

use point to calculate b:

0= -(3/4)1+b

0+3/4=b

b=3/4

tangent line is: y=-(3/4)x+3/4

 

[nycmathguy]

40.

y=x^(1/4) -x at point (1,0)

first find the slope of tangent line

(dy/dx)(x^(1/4) -x )= 1/(4x^(3/4)) - 1

compute the slope at point (1,0)

m=1/(4*1^(3/4)) - 1= -3/4

so far, tangent line is: y= -(3/4)x +b

use point to calculate b:

0= -(3/4)1+b

0+3/4=b

b=3/4

tangent line is: y=-(3/4)x+3/4
View attachment 3561

Easy enough when you provide steps as you did here.

 

[nycmathguy]

40.

y=x^(1/4) -x at point (1,0)

first find the slope of tangent line

(dy/dx)(x^(1/4) -x )= 1/(4x^(3/4)) - 1

compute the slope at point (1,0)

m=1/(4*1^(3/4)) - 1= -3/4

so far, tangent line is: y= -(3/4)x +b

use point to calculate b:

0= -(3/4)1+b

0+3/4=b

b=3/4

tangent line is: y=-(3/4)x+3/4
View attachment 3561

Can you check my work for the three unanswered calculus threads?

 

[MathLover1]

correct

 

[nycmathguy]

correct

Very nice. Better than being wrong again. See below.

Differentiate...2