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Because there are two forms of x , the form uv'+vu' must be used.

If y=sinx , dy/dx=cosx 

If y=e^kx , dy/dx=ke^kx

Therefore dy/dx=(sinx)(2e^2x)+(e^2x)(cosx)

Firstly, I would make sure that the student is aware of the basic conept of logarithims and run through the basic "laws" of logarithims to make sure that they have the knowledge they need to ans...

When you think of a number line, zero lies at the centre and the positive numbers stretch off to the right, while negative numbers go off left. This is a representation of real numbers, i...

This integral must be done using integration by parts. Therefore, we set u=ln(x) and dv=dx, which gives du=1/x and v=x.
Then, using the integration by parts formula the integral now equals x*ln(x)-i...

Two distinct real roots means that we can use b^2-4ac>0 relationship for any ax^2+bx+c equation. Apply the above gives, 4k^2 - 42(k+2)>0 Simplifying gives, k^2 - 2k -4 >0