Hi

f(x)=x+ln(x).

f '(x)=1+1/x

The equation of the tangent to the curve, at a point of absciss a is y=f '(a)(x-a)+f(a)

The only unknown is a.

Since this tangent is perpendicular to x+2y+3=0, the slopes of these two lines have to be inverse to each other (i.e. n and 1/n). The slope of the tangent is f '(a).

See here and here for similar problems.A piece of wire one meter long is to be cut into two pieces one will be used to form a square the other a circle, find the dimensions of the two that will form the least total area.

I'm not sure if this is correct but i subed x=25-.5piR into x^2+piR^2, which gave me (25-.5piR)^2 + piR^2

Your derivative is correct. Now, factorise as much as possibleDetermine the stationary points of the curve y=(x^2-1)e^-x. Using the product rule i come up with y'= -(x^2 -1)e^-x+2xe^-x , i am unsure whether i am right up to this point or if it can be simplified any further.

There is an obvious common factor in y'.