find an equation of the tangent line

Question

to the graph of y=x^4-lnx+1   the point (1,2)

Answer 1

So let's do it.

 the derivative is y' = 4x³ - 1/x

plug in the x value from the point = (1,2) and get 4(1)^3-1/1 which equals to three.
Now you can find the slope for any defined point. Using slope- intercept form

y= 3x-1

Answer 2

The slope of the tangent line is the derivative of the function evaluated at the x-coordinate of the given point (point of tangency).  

y = x⁴ - lnx + 1

 

dy/dx = 4x³ - 1/x

 

Slope of tangent line at the point (1, 2) = 4(1)³ - 1/1 = 3

 

Using the point-slope form, an equation of the tangent line is:              

     y - 2 = 3(x - 1)

 

     Simplifying, we get y = 3x - 1