A curve $R$ has equation $y=f(x)$. The point $P$ with coordinates $(7,-9)$ lies on $R$. The tangent to $R$ at point $P$ has gradient 10.

1. Find the coordinates of the corresponding point on $C$.

The translation in the negative $x$-direction by 4 units shifts the $x$-coordinate from $7$ to $7 - 4 = 3$.
The scaling parallel to the $y$-axis with a scale factor of 3 multiplies the $y$-coordinate from $-9$ to $3(-9) = -27$.
Hence, the coordinates of the corresponding point on $C$ are $(3, -27)$.

2. Determine the gradient of $C$ at this point.

A translation in the $x$-direction does not affect the gradient of the curve.
A scaling parallel to the $y$-axis by a scale factor of $3$ stretches the vertical change, multiplying the original gradient by $3$.
Gradient of $C = 3 \times 10 = 30$.

1. Find the coordinates of the corresponding point on $D$.

Since $P(7, -9)$ lies on $y = f(x)$, we have $f(7) = -9$.
For curve $D$ with equation $y = \dfrac{1}{f(x)}$, when $x = 7$,
$y = \frac{1}{f(7)} = \frac{1}{-9} = -\frac{1}{9}$
The coordinates of the point on $D$ are $\left(7, -\frac{1}{9}\right)$.

2. Differentiate $y = \dfrac{1}{f(x)}$ with respect to $x$ to find an expression for its gradient.

Writing $y = [f(x)]^{-1}$ and differentiating using the chain rule:
$\frac{dy}{dx} = -1 [f(x)]^{-2} \cdot f'(x) = -\frac{f'(x)}{[f(x)]^2}$

3. Evaluate the gradient of $D$ at $x=7$.

The gradient of the tangent to $R$ at $P$ is $10$, so $f'(7) = 10$.
Substitute $x = 7$, $f(7) = -9$, and $f'(7) = 10$ into the derivative:
$\text{Gradient} = -\frac{10}{(-9)^2} = -\frac{10}{81}$