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The easiest method to use in this incidence is integratation by parts.
So let u=x and dv/dx=exp(x). Therefore du/dx=1 and v=exp(x).
Then we use the formula where integral(udu/dx)=uv...
As this is a product of two functions it is necessary to use the product rule for differentiation. Therefore one of the functions must labeled v and the other u. i.e. u=x^3 and v=(x^2+1). It is then neces...
y = ln(2x+3 / 7x^3 +1)
d/dx(2x+3 / 7x^3 + 1) by quotient rule which is(v.du/dx - u.dv/dx) / v^2 where u=2x+3 and v=7x^3 +1 gives (-27x^3 -63x^2 +2) / (7x^3 +1)^2
so d/dx(ln(2x+3 / 7x^3 +...
Here we have a function made from the product of two functions, so we canuse the product differenciation rule.
y=uv => dy/dx=udv/dx + vdu/dx
Therefore dy/dy=(x-2)^2 + 2(x-2)(x+1)
Now although this might seem like quite a complex question at first, it's a little less intimidating when we take a while to look at it- you might notice that we can seperate x and y terms like so:1...
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