What is the equation of the line tangent to #f(x)=x^2 sin2x # at #x=pi/4#?

Question

1572 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-05T12:18:48

#y=(pix)/2-pi^2/16#

Our #x#-value = #pi/4#

To find our gradient we differentiate:
#y=f(x)g(x)# then #dy/dx=f(x)g'(x)+f'(x)g(x)#

#f(x)=x^2#
#f'(x)=2x#

#g(x)=sin(2x)#
#g'(x)=2cos(2x)#

#dy/dx=2xsin(2x)+2x^2cos(2x)#

#x=pi/4#

#dy/dx=(pi/2)sin(pi/2)+2(pi/4)^2cos(pi/2)#

#=pi/2#

#y=(pix)/2+c#

To find #y# we put #x# back into the original equation to get #y=(pi/4)^2sin(pi/2)=pi^2/16#

So, #pi^2/16=(pix)/2+c#

We put in our value for #x# and rearrange:
#pi^2/16=(pix)/2+c#

#pi^2/16=(pi(pi/4))/2+c#

#c=pi^2/16-(pi(pi/4))/2=pi^2/16-pi^2/8=-pi^2/16#

#y=(pix)/2-pi^2/16#