How do you find the equation of a line tangent to the function #y=-3/(x^2-25)# at (-4, 1/3)?
1 Answer
Aug 16, 2017
Explanation:
#•color(white)(x)m_(color(red)"tangent")=dy/dx" at x = a"#
#y=-3/(x^2-25)=-3(x^2-25)^-1#
#"differentiate using the "color(blue)"chain rule"#
#•color(white)(x)d/dx(f(g(x)))=f'(g(x))xxg'(x)#
#rArrdy/dx=3(x^2-25)^-2xx2x=(6x)/(x^2-25)^2#
#x=-4tody/dx=-24/81=-8/27#
#m=-8/27" and " (x_1,y_1)=(-4,1/3)#
#rArry-1/3=-8/27(x+4)#
#rArry=-8/27x-23/27larrcolor(red)" in slope-intercept form"#
