How do you find the equation of the line tangent to the graph of #y= ln(5 -x^2)# at the point where x = 2?

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Answer 1 · 2016-11-24T19:55:51

The equation is #y = -4x + 8#.

We start by finding the corresponding #y#-coordinate that the tangent intersects the function at.

#y = ln(5 - 2^2)#

#y = ln(1)#

#y = 0#

We now find the derivative of the function.

#y = ln(5 - x^2)#

#e^y = 5 - x^2#

#e^y(dy/dx) = 0 - 2x#

#dy/dx = -(2x)/e^y#

#dy/dx= -(2x)/e^(ln(5 - x^2))#

#dy/dx = -(2x)/(5 - x^2)#

The slope of the tangent is:

#m_"tangent" = - (2 xx 2)/(5 - 2^2)#

#m_"tangent" = -4/1#

#m_"tangent" = -4#

Hence, the equation of the tangent is:

#y - y_1 = m(x- x_1)#

#y - 0 = -4(x - 2)#

#y = -4x + 8#

Hopefully this helps!