Given that
$$y = \frac{x - 4}{2 + \sqrt{x}} \quad x > 0$$
show that
$$\frac{dy}{dx} = \frac{1}{A \sqrt{x}} \quad x > 0$$
where A is a constant to be found. - Edexcel - A-Level Maths Pure - Question 2 - 2021 - Paper 1
Question 2
Given that
$$y = \frac{x - 4}{2 + \sqrt{x}} \quad x > 0$$
show that
$$\frac{dy}{dx} = \frac{1}{A \sqrt{x}} \quad x > 0$$
where A is a constant to be found.
Worked Solution & Example Answer:Given that
$$y = \frac{x - 4}{2 + \sqrt{x}} \quad x > 0$$
show that
$$\frac{dy}{dx} = \frac{1}{A \sqrt{x}} \quad x > 0$$
where A is a constant to be found. - Edexcel - A-Level Maths Pure - Question 2 - 2021 - Paper 1
Step 1
Show that $\frac{dy}{dx}$ can be expressed using the quotient rule
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Answer
To differentiate the function, we apply the quotient rule:
dxdy=(2+x)2(2+x)dxd(x−4)−(x−4)dxd(2+x)
Calculating the derivatives:
The derivative of the numerator, dxd(x−4)=1.
The derivative of the denominator, dxd(2+x)=2x1.
Substituting these back into the formula gives:
dxdy=(2+x)2(2+x)(1)−(x−4)(2x1)
Step 2
Simplify the expression obtained for $\frac{dy}{dx}$
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Answer
Expanding and simplification yields:
dxdy=(2+x)2(2+x)−2xx−4
This leads to:
=(2+x)22+x−2xx+2x4
Combining like terms gives:
=2x(2+x)2(2x+x+4)
Step 3
Identify the constant A
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Answer
To check if the results match the form mentioned in the question, we can equate:
dxdy=Ax1
This implies that:
A=2x+x+42x(2+x)2
Thus, A must be determined from the equation derived in the previous step.
