Find the equation of the tangent line to the curve
y=(2 - (squared root x)) (1+ (square root x) +3x) at the point (1,5)

Question

Find the equation of the tangent line to the curve 
y=(2 - (squared root x)) (1+ (square root x) +3x) at the point (1,5)

campbell_st · Answer

well I'd distribute the factored form 

$$y = ( 2 - \sqrt{x})(1 + \sqrt{x} + 3x)$$

which becomes 

$$y = 2 + 2 \sqrt{x} + 6x - \sqrt{x} - x - 3x \sqrt{x}$$

collecting like terms and you get

$$y = 2 + \sqrt{x} + 5x - 3x \sqrt{x}$$

now use index notation 

$$y = 2 + x^{\frac{1}{2}} +5x -3x^{\frac{3}{2}}$$

hope that makes sense

campbell_st · Answer

next step is to fins the 1st derivative, 

this will be the equation of the slope at that point. 

Substitute x = 1 into the 1st derivative to get the slope. 

The use the point slope formula with the slope you are given... and the point (1,5)
to get the equation of the tangent

hope it helps

anonymous · Answer

I'm wondering how to do this with product rule

anonymous · Answer

just use the equation f'g + fg'