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Two functions, \(f\) and \(g\), are continuous and differentiable for all \(x\in R\). It is given that \(f(-1)=7,\ g(-1)=5\) and \(f^{′}(-1)=-4,\ g^{′}(-1)=-2\).
The gradient of the graph \(y=\dfrac{f(x)}{g(x)}\) at the point where \(x=-1\) is
\(D\)
\(\text{Using the Quotient Rule when}\ \ x=-1:\)
| \(\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)\) | \(=\dfrac{g(x)f^{′}(x)-f(x)g^{′}(x)}{g(x)^2}\) |
| \(=\dfrac{g(-1)f^{′}(-1)-f(-1)g^{′}(-1)}{g(-1)^2}\) | |
| \(=\dfrac{5 \times -4-7 \times -2}{5^2}\) | |
| \(=-\dfrac{6}{25}\) |
\(\Rightarrow D\)
It is given that \(y=f(g(x))\), where \(f(2)=5\), \(f^{′}(2)=3\), \(g(4)=2\) and \(g^{′}(4)=-2\).
What is the value of \(y^{′}\) at \(x=4\)?
\(A\)
| \(y\) | \(=f(g(x))\) | |
| \(y^{′}\) | \(=f^{′}(g(4)) \times g^{′}(4)\) | |
| \(=f^{′}(2) \times -2\) | ||
| \(=3 \times -2\) | ||
| \(=-6\) |
\(\Rightarrow A\)
Two functions, \(f\) and \(g\), are continuous and differentiable for all \(x\in R\). It is given that \(f(-2)=-7,\ g(-2)=8\) and \(f^{′}(-2)=3,\ g^{′}(-2)=2\).
The gradient of the graph \(y=f(x)\times g(x)\) at the point where \(x=-2\) is
\(D\)
\(\text{Using the Product Rule when}\ \ x=-2:\)
| \(\dfrac{d}{dx}(f(x)\times g(x))\) | \(=f(x)g^{′}(x)+g(x)f^{′}(x)\) |
| \(=f(-2)g^{′}(-2)+g(-2)f^{′}(-2)\) | |
| \(=-7\times 2+8\times 3\) | |
| \(=10\) |
\(\Rightarrow D\)
It is given that `y=f(g(x))`, where `f(1)=3`, `f^{′}(1)=-4`, `g(5)=1` and `g^{′}(5)=2`.
What is the value of `y^{′}` at `x=5`?
`A`
| `y` | `=f(g(x))` | |
| `y^{′}` | `=f^{′}(g(5)) xx g^{′}(5)` | |
| `=f^{′}(1) xx 2` | ||
| `=-4xx2` | ||
| `=-8` |
`=>A`