2015 ONE(b)

Achievement
Question
Find the gradient of the tangent to the function y=(4x−3x2)3y = (4x - 3x^2)^3 at the point (1,1)(1,1).

*You must use calculus and show any derivatives that you need to find when solving this problem.*
Official Answer
dydx=3(4x−3x2)2(4−6x)\dfrac{dy}{dx}=3(4x-3x^2)^2(4-6x)

At x=1x=1, dydx=3×1×−2=−6\dfrac{dy}{dx}=3\times1\times-2=-6
Grading Criteria

Achievement (u)

  • Correct solution with correct derivative.

Merit (r)

-

Excellence T1

-

Excellence T2

-
Exam Paper (2015)Assessment Schedule (2015)
Related Questions
2024 THREE(e)Excellence

The diagram below shows part of the graph of the function $f(x) = e^{-x^2}$, where $x \geq 0$. [DI...

2024 THREE(d)Merit

Jamie is doing some baking and pouring the flour to form a conical pile. The height o...

2024 THREE(b)Achievement

The graph below shows the function $y = f(x)$. [DIAGRAM] (i) For the function above, find the valu...

2024 TWO(e)Excellence

The graph of the function $y = \dfrac{xe^{3x}}{2x + k}$, where $k$ is a non-zero constant, has a sin...

2024 TWO(c)Merit

Show that $y = \sin(x^2) - \cos(x)$ is a solution to the equation $\dfrac{d^2y}{dx^2} + 4x^2\,y = 2...