Question

Find the derivative of
`y = ( - 2)/((x ^(2) - 3x - 4)^(2) )`

Answer 1

Derivatives

The chain rule

Given a f

unction f(x) where x is a function of t, then

f'(t)= f'(x) * x'(t)

When the function to take the derivative from is a composite function, we must use the chain rule.

Let's consider the function in the question

`y=( - 2)/((x ^(2) - 3x - 4)^(2) )`

We can manage the expression to make the function easier to take the derivative:

`y=-2\mleft(x^2-3x-4\mright)^(-2)`

The modified function can be managed as the derivative of the power function: (recall the prime sign ' means derivative).

`\mleft(x^n\mright)^(\prime)=n\cdot x^(n-1)`

But we can see the expression in parentheses is not x alone, but a function. That is why we must use the chain rule

The expression of the derivative using the chain rule is very similar:

`(g^n)^(\prime)=n\cdot g^(n-1)\cdot g^(\prime)`

Note the exponent is now applied to a base function g which derivative must be included in the final expression.

Applying the chain rule to the function:

`y^(\prime)=-2\mleft(-2\mright)\mleft(x^2-3x-4\mright)^(-3)\mleft(x^2-3x-4\mright)^(\prime)`

Note the -2 is n, the exponent is now n-1=-3

Now we have to find the remaining derivative. Let's do it apart:

`\mleft(x^2-3x-4\mright)^(\prime)=2x-3`

Now, operating in the main derivative, and substituting the above expression:

`y^(\prime)=4(x^2-3x-4)^(-3)(2x-3)`

Returning the negative exponent to the denominator:

`y^(\prime)=4((2x-3))/(\mleft(x^2-3x-4\mright)^3)`

This is the final expression for the derivative

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