Differentiation helps us determine gradients of tangents, equations of tangents and normals, stationary points, maximum or minimum values, rates of change, and small changes in quantities.
Move the value of \(x=a\) to see how the gradient of the tangent to \(f(x)=x^2\) changes at different points.
The values match those in the textbook example: −2, −1, 0, 1 and 2.
The tangent line is horizontal.
Negative gradient — the tangent line slants to the left.
Zero gradient — the tangent line is horizontal.
Positive gradient — the tangent line slants to the right.
Connecting three rates of change: \(\dfrac{dy}{dx}\), \(\dfrac{dx}{dt}\) and \(\dfrac{dy}{dt}\).
The value of \(y\) is increasing with respect to time.
The value of \(y\) is decreasing with respect to time.
Use the first and second order derivatives according to the type of problem.
For \(y=f(x)\), the gradient of the tangent at point \(P(a,f(a))\) is \(f'(a)\).
Tangent: \(y-f(a)=f'(a)(x-a)\). If \(f'(a)\neq0\), the normal is: \(y-f(a)=-\dfrac{1}{f'(a)}(x-a)\).
A stationary point occurs when \(\dfrac{dy}{dx}=0\). The tangent to the graph at that point is horizontal.
A sign change of \(+\to-\) indicates a maximum point; \(-\to+\) indicates a minimum point; no sign change indicates a point of inflection.
When \(\dfrac{dy}{dx}=0\): if \(\dfrac{d^2y}{dx^2}<0\), the turning point is a maximum; if \(\dfrac{d^2y}{dx^2}>0\), it is a minimum.
Express the function in one variable, find the first derivative, set it equal to zero, and determine the nature of the stationary point obtained.
Use: \(\dfrac{dy}{dt}=\dfrac{dy}{dx}\times\dfrac{dx}{dt}\). A negative value shows that the quantity is decreasing.
If \(\delta x\) is small: \(\delta y\approx\dfrac{dy}{dx}\delta x\) and \(f(x+\delta x)\approx f(x)+\dfrac{dy}{dx}\delta x\).
Identify the quantity to be maximised or minimised.
Express the function in one variable only.
Find the first derivative and set it equal to zero.
Determine the nature of the point and interpret the answer in context.
Small change in \(y\)
The smaller \(\delta x\) is, the more accurate.
Approximate value of function
Use this to estimate a new value of \(y\).
Answer in your mind, then press "Check Answer".
For a curve, the value of \(\dfrac{dy}{dx}\) at point \(P\) is \(-3\). What is the condition of the tangent line?
As \(x\) increases through a stationary point, the sign of \(\dfrac{dy}{dx}\) changes from positive to negative. What is the nature of that stationary point?
Given \(\dfrac{dy}{dx}=4\) and \(\dfrac{dx}{dt}=3\), find \(\dfrac{dy}{dt}\).
If \(\dfrac{dy}{dx}=0\), the point must be a maximum point.
The point may be a maximum, a minimum, or a point of inflection — the sign change of \(\dfrac{dy}{dx}\) must be checked.
If \(\dfrac{d^2y}{dx^2}=0\), the point must be a point of inflection.
The sign change of \(\dfrac{dy}{dx}\) must still be checked using the tangent sketch method to determine the nature of the stationary point.
A negative rate of change means the calculation is wrong.
A negative sign for a rate of change simply means the quantity is decreasing. It is mathematically correct.
Objective: Identify gradients, stationary points, the chain rule and key terms in applications of differentiation.
1 If \(f'(a) < 0\), what is the condition of the tangent line at that point?
2 The sign of \(\dfrac{dy}{dx}\) changes from positive to negative as \(x\) increases. What is the nature of the stationary point?
3 A point where \(\dfrac{dy}{dx}=0\) is called a ________ point.
4 The relation \(\dfrac{dy}{dt}=\dfrac{dy}{dx}\times\dfrac{dx}{dt}\) is called the ________.
Drag each term to the correct meaning. On a phone, tap a term first, then tap the matching box.
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