• The gradient changes at different points on a curve.
• The gradient function \(f'(x)\) is used to determine the gradient of tangent to the curve at any point on the function graph \(f(x)\).
• For example, for the function \(f(x)=x^2\):

• Based on the above diagram, the formula for the equation of straight line \(l\) with gradient \(m\) that passes through point \(P(x_1, y_1)\) can be written as:

\(y-y_1=m(x-x_1)\)

• The equation of the tangent is:

\(y-f(a)=f'(a)(x-a)\)

• The line which is perpendicular to the tangent is the normal to the curve  \(y=f(x)\) at point \(P(a,f(a))\).

•   If the gradient of the tangent, \(f'(a)\) exists and is non-zero, the gradient of the normal based on the relation of \(m_1m_2=-1\) is:

\(-\dfrac{1}{f'(a)}\)

• The equation of the normal is:

\(y-f(a)=-\dfrac{1}{f'(a)}(x-a)\)

• Turning points.
• When \(x\) increases through \(x=c\), the value of \(\dfrac{dy}{dx}\) changes sign from positive to negative.

• Turning points.
• When \(x\) increases through \(x=a\), the value of \(\dfrac{dy}{dx}\) changes sign from negative to positive.

• This stationary point which is not a maximum or a minimum point.
• A point on the curve at which the curvature of the graph changes.
• When \(x\) increases through \(x=b\), the value of \(\dfrac{dy}{dx}\) does not change in sign.

A turning point on a curve \(y=f(x)\) is a maximum point when \(\dfrac{dy}{dx}=0\) and \(\dfrac{d^2y}{dx^2} \lt 0\).

A turning point on a curve \(y=f(x)\) is a minimum point when \(\dfrac{dy}{dx}=0\) and \(\dfrac{d^2y}{dx^2} \gt 0\).

By applying the chain rule concept, if two variables, \(y\) and \(x\) change with time, \(t\) and are related by the equation \(y=f(x)\), then the rates of change \(\dfrac{dy}{dt}\) and \(\dfrac{dx}{dt}\) can be related by:

\(\dfrac{dy}{dt}=\dfrac{dy}{dx}\times \dfrac{dx}{dt}\)

• If \(y=f(x)\) and the small change in \(x\), that is \(\delta x\) causes a small change in \(y\), \(\delta y\), then:

\(\delta y\approx\dfrac{dy}{dx} \times \delta x\)

• Since \(f(x+\delta x)=y+\delta y\) and \(\delta y\approx\dfrac{dy}{dx} \times \delta x\), then:

\(f(x+\delta x)=y+\dfrac{dy}{dx}\delta x\) or \(f(x+\delta x)=f(x)+\dfrac{dy}{dx}\delta x\)

• If \(x\) changes from \(x\) to \(x+\delta x\), then:
  • The percentage change in \(x=\dfrac{\delta x}{x}\times 100\%\)
  • The percentage change in \(y=\dfrac{\delta y}{y}\times 100\%\)

Find the equation of the tangent and normal to the curve \(f(x)=x^3-2x^2+5\) at point \(P(2,5)\).

Given,

\(f(x)=x^3-2x^2+5\).

So,

\(f'(x)=3x^2-4x\).

At point \(P(2,5)\),

Gradient of the tangent:

\(\begin{aligned} f'(2)&=3(2)^2-4(2) \\ &=3(4)-8 \\ &=4. \end{aligned}\)

Equation of the tangent:

\(\begin{aligned} y-5&=4(x-2) \\ y-5&=4x-8 \\ y&=4x-8+5 \\ y&=4x-3. \end{aligned}\)

Gradient of the normal at point \(P(2,5)\):

\(\begin{aligned} m_Tm_N&=-1 \\ (4)m_N&=-1 \\ m_N&=-\dfrac{1}{4}. \end{aligned}\)

Equation of the normal:

\(\begin{aligned} y-5&=-\dfrac{1}{4}(x-2) \\ y-5&=-\dfrac{1}{4}x+\dfrac{1}{2} \\ y&=-\dfrac{1}{4}x+\dfrac{1}{2}+5 \\ y&=-\dfrac{1}{4}x+\dfrac{11}{2}. \end{aligned}\)

Given \(y=x^3\), find 

(a) the approximate change in \(y\) when \(x\) increases from \(4\) to \(4.05\),
(b) the approximate change in \(x\) when \(y\) decreases from \(8\) to \(7.97\).

(a)

\(y=x^3\),

\(\dfrac{dy}{dx}=3x^2\).

When \(x=4\),

\(\begin{aligned} \delta x&=4.05-4 \\ &=0.05 \\\\ \dfrac{dy}{dx}&=3(4)^2 \\ &=48 \\\\ \delta y &\approx \dfrac{dy}{dx}\times \delta x \\ &=48 \times 0.05 \\ \delta y &=2.4. \end{aligned}\)

Therefore, the approximate change in \(y\), \(\delta y\) is \(2.4\).

\(\delta y \gt 0\) means there is a small increase in \(y\) of \(2.4\).

(b)

When \(y=8\),

\(\begin{aligned} x^3&=8 \\ x&=2 \\\\ \delta y&=7.97-8 \\ &=-0.03 \\\\ \dfrac{dy}{dx}&=3(2)^2 \\ &=12 \\\\ \delta y &\approx \dfrac{dy}{dx} \times \delta x \\ -0.03&=12\times\delta x \\ \delta x&=\dfrac{-0.03}{12} \\ \delta x&=-0.0025. \end{aligned}\)

Therefore, the approximate change in \(x\), \(\delta x\) is \(-0.0025\).

\(\delta x \lt 0\) means there is a small decrease in \(x\) of \(0.0025\).

• The gradient changes at different points on a curve.
• The gradient function \(f'(x)\) is used to determine the gradient of tangent to the curve at any point on the function graph \(f(x)\).
• For example, for the function \(f(x)=x^2\):

• Based on the above diagram, the formula for the equation of straight line \(l\) with gradient \(m\) that passes through point \(P(x_1, y_1)\) can be written as:

\(y-y_1=m(x-x_1)\)

• The equation of the tangent is:

\(y-f(a)=f'(a)(x-a)\)

• The line which is perpendicular to the tangent is the normal to the curve  \(y=f(x)\) at point \(P(a,f(a))\).

•   If the gradient of the tangent, \(f'(a)\) exists and is non-zero, the gradient of the normal based on the relation of \(m_1m_2=-1\) is:

\(-\dfrac{1}{f'(a)}\)

• The equation of the normal is:

\(y-f(a)=-\dfrac{1}{f'(a)}(x-a)\)

• Turning points.
• When \(x\) increases through \(x=c\), the value of \(\dfrac{dy}{dx}\) changes sign from positive to negative.

• Turning points.
• When \(x\) increases through \(x=a\), the value of \(\dfrac{dy}{dx}\) changes sign from negative to positive.

• This stationary point which is not a maximum or a minimum point.
• A point on the curve at which the curvature of the graph changes.
• When \(x\) increases through \(x=b\), the value of \(\dfrac{dy}{dx}\) does not change in sign.

A turning point on a curve \(y=f(x)\) is a maximum point when \(\dfrac{dy}{dx}=0\) and \(\dfrac{d^2y}{dx^2} \lt 0\).

A turning point on a curve \(y=f(x)\) is a minimum point when \(\dfrac{dy}{dx}=0\) and \(\dfrac{d^2y}{dx^2} \gt 0\).

By applying the chain rule concept, if two variables, \(y\) and \(x\) change with time, \(t\) and are related by the equation \(y=f(x)\), then the rates of change \(\dfrac{dy}{dt}\) and \(\dfrac{dx}{dt}\) can be related by:

\(\dfrac{dy}{dt}=\dfrac{dy}{dx}\times \dfrac{dx}{dt}\)

• If \(y=f(x)\) and the small change in \(x\), that is \(\delta x\) causes a small change in \(y\), \(\delta y\), then:

\(\delta y\approx\dfrac{dy}{dx} \times \delta x\)

• Since \(f(x+\delta x)=y+\delta y\) and \(\delta y\approx\dfrac{dy}{dx} \times \delta x\), then:

\(f(x+\delta x)=y+\dfrac{dy}{dx}\delta x\) or \(f(x+\delta x)=f(x)+\dfrac{dy}{dx}\delta x\)

• If \(x\) changes from \(x\) to \(x+\delta x\), then:
  • The percentage change in \(x=\dfrac{\delta x}{x}\times 100\%\)
  • The percentage change in \(y=\dfrac{\delta y}{y}\times 100\%\)

Find the equation of the tangent and normal to the curve \(f(x)=x^3-2x^2+5\) at point \(P(2,5)\).

Given,

\(f(x)=x^3-2x^2+5\).

So,

\(f'(x)=3x^2-4x\).

At point \(P(2,5)\),

Gradient of the tangent:

\(\begin{aligned} f'(2)&=3(2)^2-4(2) \\ &=3(4)-8 \\ &=4. \end{aligned}\)

Equation of the tangent:

\(\begin{aligned} y-5&=4(x-2) \\ y-5&=4x-8 \\ y&=4x-8+5 \\ y&=4x-3. \end{aligned}\)

Gradient of the normal at point \(P(2,5)\):

\(\begin{aligned} m_Tm_N&=-1 \\ (4)m_N&=-1 \\ m_N&=-\dfrac{1}{4}. \end{aligned}\)

Equation of the normal:

\(\begin{aligned} y-5&=-\dfrac{1}{4}(x-2) \\ y-5&=-\dfrac{1}{4}x+\dfrac{1}{2} \\ y&=-\dfrac{1}{4}x+\dfrac{1}{2}+5 \\ y&=-\dfrac{1}{4}x+\dfrac{11}{2}. \end{aligned}\)

Given \(y=x^3\), find 

(a) the approximate change in \(y\) when \(x\) increases from \(4\) to \(4.05\),
(b) the approximate change in \(x\) when \(y\) decreases from \(8\) to \(7.97\).

(a)

\(y=x^3\),

\(\dfrac{dy}{dx}=3x^2\).

When \(x=4\),

\(\begin{aligned} \delta x&=4.05-4 \\ &=0.05 \\\\ \dfrac{dy}{dx}&=3(4)^2 \\ &=48 \\\\ \delta y &\approx \dfrac{dy}{dx}\times \delta x \\ &=48 \times 0.05 \\ \delta y &=2.4. \end{aligned}\)

Therefore, the approximate change in \(y\), \(\delta y\) is \(2.4\).

\(\delta y \gt 0\) means there is a small increase in \(y\) of \(2.4\).

(b)

When \(y=8\),

\(\begin{aligned} x^3&=8 \\ x&=2 \\\\ \delta y&=7.97-8 \\ &=-0.03 \\\\ \dfrac{dy}{dx}&=3(2)^2 \\ &=12 \\\\ \delta y &\approx \dfrac{dy}{dx} \times \delta x \\ -0.03&=12\times\delta x \\ \delta x&=\dfrac{-0.03}{12} \\ \delta x&=-0.0025. \end{aligned}\)

Therefore, the approximate change in \(x\), \(\delta x\) is \(-0.0025\).

\(\delta x \lt 0\) means there is a small decrease in \(x\) of \(0.0025\).