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We are given the substitution to use, so the first step is to differentiate "u" with respect to x. 

du/dx = -sin(x)

Now, to replace the "dx&q...

As we are differentiating a product (two things times together) we can use the product rule which is if:

                       y = u(x)v(x)

then

             ...

When we integrate a function we must first raise the power of x in each term by one. The first term therefore becomes x^(5/2). The second term can be thought of as x^0 which we know that any number to ...

To solve this problem, you need to put it into simplest form which is putting it into natural logarithm both the RHS and LHS. Then differentiate both side with respect to x as shown below.

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As the name suggests, the product rule is used to differentiate a function in which a product of 2 expressions in x exists. This means the two expressions in x are multiplied by each other, even when t...