Calculus help please?

Question

5x^3+radicalx (1,6) How do I write an equation for this? (tangent line)

Answer 1

The function is f(x) = 5x³ + √x .

Slope of the tangent line is f '(x) at x = 1 .

f '(x) = 5*3x² + 1/2√x                     [ derivative of  xⁿ = n x^n-1 and deriavtive of √x is  1/2√x  ]        

f '(x) = 15x² + 1/2√x  

f '(1) = 15(1)² + 1/2√1 

f '(1) = 15 + 1/2

f '(1) = 31/2 

So the slope of tangent is 31/2 .

The equation of tangent with slope 31/2  and point (1,6) using slope point form .

y – y₁ = m(x – x₁)

y – 6 = 31/2 (x – 1)

2(y - 6) = 31(x – 1)

2y -12 = 31x -31

31x -2y -19 = 0

So the equation of tangent is 31x -2y -19 = 0 .