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astarmathsandphysics:
Some more doubts...

1. Consider the trigonometric curve y=sin(2x-pi/2)
a) Find dy/dx and d2y/dx2

2. Find an equation for a line that is tangent to the graph of y=e^x that passes through the origin

3. Find the derivatice of y with repect to x, dy/dx by implicity differentiation: xy(x+y)^1/2=1

4. Find the derivative of y with repect to x, dy/dx: ln(1+x^2)^1/2=xarctanx

Thanks!
a)dy/dx=2cos(2x-pi/2) and d2y/d2x =-4sin(2x-pi/2)
b)dy/dx =e^x so at x=0 gradient is e^0 =1 and y=1
y-1=1(x-0) so y=x+1
c)o sh*t
y(x+y)^1/2 +x(x+y)^1/2 dy/dx +1/2xy(x+y)^-1/2 (1+dy/dx) =0
dy/dx(x(x+y)^1/2+1/2xy(x+y)^-1/2 )=-y(x+y)^1/2 -1/2xy(x+y)^-1/2
dy/dx =(-y(x+y)^1/2 -1/2xy(x+y)^-1/2)/(x(x+y)^1/2+1/2xy(x+y)^-1/2 ) =(-y(x+y)-1/2xy)/(x(x+y)+1/2xy) =(-y^2-3/2xy)/(x^2 +3/2xy)
4. Where is y here

IGSTUDENT:
thanks for the help

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