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Find the coordinates of the point on the curve
y=(2(x-5))/(sqrt(x+1)) where the gradient is
(5)/(4).

Find the coordinates of the point on the curve $\newline$ $y=\frac{2(x-5)}{\sqrt{x+1}}$ where the gradient is $\newline$ $\frac{5}{4}$ .

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Find derivatives using logarithmic differentiation

Full solution

Q. Find the coordinates of the point on the curve

\newline

y=\frac{2(x-5)}{\sqrt{x+1}}

where the gradient is

\newline

\frac{5}{4}

.

Find Derivative: First, we need to find the derivative of $y$ with respect to $x$ to determine the gradient of the curve. The function is $y = \frac{2(x-5)}{\sqrt{x+1}}$ . Using the quotient rule, the derivative $y'$ is given by: $y' = \frac{[\sqrt{x+1} \cdot 2] - [2(x-5) \cdot (\frac{1}{2})(x+1)^{-\frac{1}{2}}]}{x+1}$ = $\frac{2\sqrt{x+1} - \frac{x-5}{\sqrt{x+1}}}{x+1}$
Set Equal to $\frac{5}{4}$ : Next, set the derivative equal to $\frac{5}{4}$ to find the x-coordinate where the gradient is $\frac{5}{4}$ .
$\frac{2\sqrt{x+1} - \frac{x-5}{\sqrt{x+1}}}{x+1} = \frac{5}{4}$
Cross-multiplying to clear the fraction:
$4(2\sqrt{x+1} - \frac{x-5}{\sqrt{x+1}}) = 5(x+1)$
$8\sqrt{x+1} - \frac{4(x-5)}{\sqrt{x+1}} = 5x + 5$
Simplify and Solve: Simplify and solve for $x$ : $8\sqrt{x+1} - \frac{4(x-5)}{\sqrt{x+1}} = 5x + 5$ Let's multiply through by $\sqrt{x+1}$ to clear the square root: $8(x+1) - 4(x-5) = 5x + 5 \sqrt{x+1}$ $8x + 8 - 4x + 20 = 5x + 5 \sqrt{x+1}$ $4x + 28 = 5x + 5 \sqrt{x+1}$

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