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Differentiate y=x*ln(x^3-5)

We can immediately see that more than differentiation rule will be needed here. The expression in question is the product of two smaller expressions, so the product rule may be useful. But to apply the product rule, we need to be able to differentiate the two smaller expressions. ln(x3-5) is slightly more complicated to differentiate. However, notice it is the composition of two functions we know how to differentiate: x3-5 and ln(x). This suggest we may be able to apply the chain rule.

First, let u=ln(x3-5)

and v=x3-5

Then u=ln(v)

Differentiating u and v:

du/dv=1/v

dv/dx=3x2

Recall the formula for the chain rule, which in this case is du/dx=(du/dv)*(dv/dx)

Substituting into the chain rule:

du/dx=(du/dv)*(dv/dx)

=(1/v)*(3x2)

=3x2/v

=3x2/(x3-5)

So, d/dx(ln(x3-5))=3x2/(x3-5)

Note – In an exam, it may be faster simply to use the standard formula for differentiating ln: d/dx(ln(f(x)))=f'(x)/f(x) . You can use this formula whenever you spot you are differentiating ln of some function. You should be able to see how this would work in the above example. I have provided a full method for clarity, not because it is necessary to do so in your exam.

We are now in a position to apply the product rule. Recall that the formula for the product rule is d/dx(UV)=V*dU/dx+U*dV/dx   (U and V here used just to avoid confusion with u and v used earlier)

Let U=x

and V=ln(x3-5)

then dU/dx=1

and dV/dx=3x2/(x3-5)

Substituting into the product rule formula:

d/dx(x*ln(x3-5))=dx(UV)

=V*dU/dx+U*dV/dx

=ln(x3-5)*1+x*3x2/(x3-5)

=ln(x3-5)+3x3/(x3-5)

This gives us our answer:

dy/dx=ln(x3-5)+3x3/(x3-5)

Nat N. GCSE Maths tutor, A Level Maths tutor, GCSE Further Mathematic...

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