Q10. Please answer all the parts to this question The graphs of the functions \( f \) and \( g \) are shown in the figure.**Graph Description:**- The blue graph represents the function \( f \).- The green graph represents the function \( g \).- The \( x \)-axis ranges from \(-1\) to 6.- The \( y \)-axis ranges from 0 to 4.**Function \( f \):**- Starts at \( (-1, 0.5) \), rises to \( (1, 2) \), peaks at \( (3, 4) \), and then descends to \( (6, 2) \).**Function \( g \):**- Begins at \( (0, 4) \), decreasing linearly to \( (3, 1) \), and then increases to \( (6, 3) \).**Mathematical Expressions:**Let \( u(x) = f(x)g(x) \) and \( v(x) = \frac{f(x)}{g(x)} \).**Tasks:**(a) Find \( u'(1) \).(b) Find \( v'(5) \).

Given that: ux=fxgxand vx=fxgxThe Blue line represents the graph of fx and green line represents the graph ofgx Since both lines are meeting twice to each other, we will get two values.for blue line fx that passes 0,0and 2,4.let, x1,y1=0,0 and x2,y2=2,4equation of line will be,y-y1=y2-y1x2-x1x-x1y-0=4-02-0x-0fx=y=2xIn the same way,Green line gx passes points0,2and 2,0compare withx1,y1 and x2,y2Equation of line:y-y1=y2-y1x2-x1x-x1y-2=0-22-0x-0y-2=-xgx=y=2-xNow,ux=fxgxux=2x2-xux=4x-2x2Differentiate with respect to x,u'x=4-4xput x=1u'1=4-41=4-4=0u'(1)=0vx=fxgx=2x2-xdifferentiate with respect to x,v'x=2-xddx2x-2xddx2-x2-x2=22-x-2x-12-x2=4-2x+2x2-x2v'x=42-x2put x=5,v'5=52-52=59v'(5)=59Now, we will find for furthur part,fx has points2,4 and5,3Equation of line,y-4=3-45-2x-23y-12=-x+2fx=y=-x+143For green line, which passes 2,0and 5,2Equation of line is:y-0=2-05-2x-2g(x)=y=2x-23again, ux=fxgx=-x+143×2x-23=2-x2+2x+14x-289ux=29-x2+16x-28Differentiate with respect to x,u'x=29-2x+16Put x=1u'1=29×14=289u'(1)=289Again, vx=fxgx=-x+1432x-23=-x+142x-2differentiate with respect to x,v'x=12x-2ddx-x+14--x+14ddxx-2x-22v'x=12x-2-1--x+141x-22Put x=5v'5=12-5-2--5+145-22=12-3-99=-23v'(5)=-23