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LHS: because alnx = lnx^a, 2ln2x = ln(2x)² = ln4x²

Now, because ln and e are inverse functions, we take both sides to the power of e. Therefore:

e^{ln4x^2 ...}

We must first use trigonometric identities to simplify cos²(x). We can use the formula cos(A+B) = cos(A)cos(B) - sin(A)sin(B) , where A=x and B=x, so that we ...

First we find the y coordinate which is a function of x:

x = 2+ h so  y = (2 - 2 - h)(1 + 2 + h) + 3 = -h² - 3h + 3

Now for the gradient, the line passes through points (2,3) and ...

Consider the function as 1(ln(x)) and then use integrating by parts, where you would differentiate the ln(x) part.