STEP II 2007 question 6 solution - The Student Room
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STEP II 2007 question 6 solution

Part 1

\frac{d}{dx} ln (u) = \frac{u'}{u}

\mathrm{Let} u = x + (3+x^2)^\frac{1}{2}

\frac{d}{dx} u = 1 + \frac{1}{2}(2x)(3+x^2)^\frac{-1}{2}

 = 1 + \frac{x}{(3+x^2)^\frac{1}{2}}

We can write this as

 \frac{\sqrt{(3+x^2)}}{\sqrt{(3+x^2)}} + \frac{x}{\sqrt{3+x^2}}

 = \frac{x + \sqrt{(3+x^2)}}{\sqrt{(3+x^2)}}

Going back we have

 = \frac{x + \sqrt{(3+x^2)}}{\sqrt{(3+x^2)}} \times \frac{1}{x + \sqrt {3+x^2}}

Which cancels down to

 \frac {1}{\sqrt{3+x^2}}


For the next bit

 \frac{d}{dx} x(\sqrt{3+x^2})


We can use the product rule... differentiate*leave + leave*differentiate (my way of remembering it)

So we have,

 1 \cdot \sqrt{3+x^2} + x \cdot \frac{x}{\sqrt{3+x^2}}

 = \frac {\sqrt{3+x^2} \cdot \sqrt{3+x^2}}{\sqrt{3+x^2}} + \frac{x^2}{\sqrt{3+x^2}}

Which simplifies to

 = \frac {2x^2+3}{\sqrt{3+x^2}}




In progress

Solution by dramaminedreams

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