Can someone me on question 9 & 10. one is just fine. you

9. y = x - 1/2 + π/4

10. y = (-2/3)(√6)x + (2/3)√3 + π/3

Step-by-step explanation:

9. From a table of derivatives,

... arctan'(x) = 1/(1+x²)

Then, using the chain rule, ...

... f'(x) = arctan'(2x) = 2/(1 +(2x)²) = 2/(1 +4x²)

For x=1/2, the slope of the tangent is ...

... f'(1/2) = 2/(1 +4·(1/2)²) = 1

Then in point-slope form, the equation of the tangent line is

... y = 1(x -1/2) +π/4

... y = x -1/2 +π/4

10. From a table of derivatives,

... arccos'(x) = -1/√(1 -x²)

Then, using the chain rule, ...

... arccos'(x²) = -2x/√(1 -x⁴)

For x=1/√2, the slope of the tangent is ...

... f'(1/√2) = -(2/√2)/√(1 -(1/√2)⁴) = (-√2)/√(3/4) = -2√(2/3)

... f'(1/√2) = (-2/3)√6

Then in point-slope form, the equation of the tangent line is

... y = (-2/3)(√6)(x -1/√2) +π/3

... y = (-2/3)(√6)x +(2/3)√3 +π/3