Recent content by Calc.Hatr10

Find tangent line to y= sqrt(25-x^2) at the point (4,3) f'(x)= (25-x^2)^1/2 f'(x)= 1/2(25-x^2)^-1/2 f'(x)= (-2x)1/2(25-x^2)^-1/2 f'(x)= -x(25-x^2)^-1/2 f'(4)=-4(25-4^2)^-1/2 f'(4)=-4(25-(4^2))^-1/2 f'(4)=-4(25-16)^-1/2 f'(4)=-4(9)^-1/2 f'(4)=-4(1/3) f'(4)=-4/3 3=-4/3(4)+b...

so.. f'(4)=4(25-(4^2))^-1/2 f'(4)=4(25-16)^-1/2 f'(4)=4(9)^-1/2 f'(4)=4(1/3) f'(4)=4/3 ?

Im checking over my work thus far and i cannot see where I lost a neg. at (25-4^2) -4^2=+16 Please explain

the point is (4,3) f'(x)= x(25-x^2)^-1/2 f'(4)= 4(25-4^2)^-1/2 f'(4)= 4(25+16)^-1/2 f'(4)= 4(41)^-1/2

yes I tried the chain rule...and I can't seem to find the slope

Homework Statement Find the equation of the tangent line to y=sqrt(25-x^2) at the point (4,3) Homework Equations The Attempt at a Solution (25-x^2)^1/2 1/2(25-x^2)^-1/2 (2x)1/2(25-x^2)^-1/2 x(25-x^2)^-1/2 I have no idea what to do or where to go after this... Please Help!