How do you find the equation of the tangent line to graph #y=x^sinx# at the point #(pi/2, pi/2)#?

1 Answer
Nov 3, 2016

The equation of the tangent line is #y= x#.

Explanation:

#lny = ln(x^sinx)#

#lny= sinxlnx#

Now, use the product rule and implicit differentiation to find the derivative.

#1/y(dy/dx) = (cosx xx lnx + sinx xx 1/x)#

#1/y(dy/dx) = cosxlnx + sinx/x#

#dy/dx = (cosxlnx + sinx/x)/(1/y)#

#dy/dx = y(cosxlnx + sinx/x)#

Since #y = x^sinx#:

#dy/dx = x^sinx(cosxlnx + sinx/x)#

All that is left to do is find the slope of the tangent. This can be found by inserting #x = a# into the derivative, since the derivative represents the instantaneous rate of change of the function.

#m_"tangent" = (pi/2)^(sin(pi/2))(cos(pi/2)ln(pi/2) + (sin(pi/2))/(pi/2))#

#m_"tangent" = pi/2(0 + 1/(pi/2))#

#m_"tangent" = pi/2(0 + 2/pi)#

#m_"tangent" = 1#

Now, we use point-slope form to find the equation of the line.

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

#y- pi/2 = 1(x- pi/2)#

#y - pi/2 = x - pi/2#

#y = x - pi/2 + pi/2#

#y = x#

Hopefully this helps!