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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
\(D\)
\(\text{Using the Product Rule when}\ \ x=-2:\)
| \(\dfrac{d}{dx}(f(x)\times g(x))\) | \(=f(x)g^{′}(x)+g(x)f^{′}(x)\) |
| \(=f(-2)g^{′}(-2)+g(-2)f^{′}(-2)\) | |
| \(=-7\times 2+8\times 3\) | |
| \(=10\) |
\(\Rightarrow D\)
Evaluate `f^{′}(4)`, where `f(x) = xsqrt(2x + 1)`. (3 marks)
`13/3`
`f(x) = xsqrt(2x + 1)`
| `f^{′}(x)` | `=1 sqrt(2x + 1) + x xx 1/2 xx 2(2x+1)^(-1/2)` | |
| `=sqrt(2x + 1) + x(2x+1)^(-1/2)` | ||
| `f^{′}(4)` | `= sqrt9 + 4(9)^(-1/2)` | |
| `=3 + 4/3` | ||
| `=13/3` |
Let `g(x) = (2-x^3)^3`.
Evaluate `g^{′}(-1)`. (2 marks)
`-81`
`text(Using Chain Rule:)`
| `g^{′}(x)` | `= 3 (2-x^3)^2 (-3x^2)` |
| `= -9x^2 (2-x^3)^2` | |
| `:. g^{′}(1)` | `= -9 (–1)^2 [2-(–1)^3]^2` |
| `= -81` |