f(x)=(3x^2-2)/(6x)f(x)=3x2−26x
Let us first find the slope of the tangent.
Slope of the tangent at a point is the first derivative of the curve at the point.
so First derivative of f(x) at x=1 is the slope of the tangent at x=1
To find f'(x) we need to use quotient rule
Quotient rule: d/dx(u/v)=((du)/dxv-u(dv)/dx)/v^2ddx(uv)=dudxv−udvdxv2
u=3x^2-2=>(du)/dx=6xu=3x2−2⇒dudx=6x
v=6x=>(dv)/dx=6v=6x⇒dvdx=6
f'(x)=((du)/dxv-u(dv)/dx)/v^2
f'(x)=(6x(6x)-(3x^2-2)6)/(6x)^2
f'(x)=(36x^2-18x^2+12)/(6x)^2color(blue) "combine the like terms"
f'(x)=(18x^2+12)/(36x^2)color(blue)"factor out 6 on the numerator "
f'(x)=(6(3x^2+2))/(36x^2)color(blue)"cancel the 6 with the 36 in the denominator"
f'(x)=(3x^2+2)/(6x^2)
f'(1)=(3+2)/6=>f'(1)=5/6
color(green)"slope of the tangent=5/6"
color(green)"slope of the normal=negative reciprocal of slope of the tangent=-6/5"
f(1)=(3-2)/6=>f'(1)=1/6
color( red)" point-slope form of an equation of line "
color( red) "y-y1=m(x-x1)...(where m:slope,(x1,y1) :points)"
We have slope =-6/5 and the points are (1,1/6)
Use the point slope form
y-(1/6)=-6/5(x-1)=>y=(-6/5)x+6/5+1/6
color(green)"combine the constant terms"
color(green)"y=-6/5x+41/30"
