MA 1505 Tutorial 1: Derivative

Definition of Derivative:

$f^{'}(x)=\lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$

Rule: Assume f(x) and g(x) are two differentiable functions, the basic rules of derivative are

$(f\pm g)^{'}(x)=f^{'}(x)\pm g^{'}(x)$

$(f\cdot g)^{'}(x)= f^{'}(x) g(x) + f(x)g^{'}(x)$

$(f/g)^{'}(x)=(f^{'}(x)g(x)-f(x)g^{'}(x))/(g(x))^{2}$

$(f\circ g)^{'}(x)=f^{'}(g(x))g^{'}(x)$

Definition of Critical Point: $x_{0}$ is called a critical point of f(x), if $f^{'}(x_{0})=0.$

If $f^{'}(x)>0$ on some interval I, then f(x) is increasing on the interval I. Similarly, if $f^{'}(x)<0$ on some interval I, then f(x) is decreasing on the interval I.

Tangent Line: Assume f(x) is a differentiable function on the interval I, then the tangent line of f(x) at the point $x_{0}\in I$ is $y-f(x_{0})=f^{'}(x_{0})(x-x_{0}),$ where $f^{'}(x_{0})$ is the slope of the tangent line.

Derivative of Parameter Functions: Assume y=y(t) and x=x(t), the derivative $y^{'}(x)$ is $y^{'}(t)/x^{'}(t),$ because the Chain Rule of derivatives.

Question 1. Calculate the tangent line of the curve $x^{\frac{1}{4}} + y^{\frac{1}{4}}=4$ at the point (16,16).

Method (i). Take the derivative of the equation $x^{\frac{1}{4}}+y^{\frac{1}{4}}=4$ at the both sides, we get

$\frac{1}{4}x^{-\frac{3}{4}} + \frac{1}{4}y^{-\frac{3}{4}} y^{'}=0.$

Assume x=y=16, we have the derivative $y^{'}(16)=-1.$ That means the tangent line of the curve at the point (16,16) is y-16=-(x-16). i.e. y=-x+32.

Method (ii). From the equation, we know $y(x)=(4-x^{\frac{1}{4}})^{4}$, then calculating the derivative directly. i.e.

$y^{'}(x)=4(4-x^{\frac{1}{4}})^{3}\cdot (-1)\cdot \frac{1}{4}x^{-\frac{3}{4}}$

Therefore, $y^{'}(16)=-1.$

Method (iii). Making the substitution $x=4^{4}\cos^{8}\theta, y=4^{4}\sin^{8}\theta,$ then (16,16) corresponds to $\theta=\pi/4.$ From the derivative of the parameter functions, we know

$\frac{dy}{dx}= \frac{dy/d\theta}{dx/d\theta}=\frac{4^{4}\cdot 8\sin^{7}\theta\cdot \cos\theta}{4^{4}\cdot 8\cos^{7}\theta\cdot (-\sin\theta)}$

If we assume $\theta=\pi/4,$ then $y^{'}(16)=-1.$

Method (iv). Geometric Intuition. Since the equation $x^{\frac{1}{4}}+y^{\frac{1}{4}}=4$ is a symmetric graph with the line y=x, and (16,16) is also on the symmetric line. Therefore, the slope of the curve at the point (16,16) is -1. Hence, the tangent line is y=-x+32.

Question 2. Let $y=(1+x^{2})^{-2}$ and $x=\cot \theta.$ Find dy/dx and express your answer in terms of $\theta.$

Method (i). $y=\frac{1}{1+x^{2}}= \sin^{2}\theta$

$\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta} = \frac{2\sin\theta \cos \theta}{-\sin^{-2}\theta}= - \sin^{2}\theta\sin2\theta.$

Method (ii). $\frac{dy}{dx}=-\frac{2x}{(1+x^{2})^{2}} = -\frac{2\cot \theta}{(1+\cot^{2}\theta)^{2}}=-\sin^{2}\theta\sin 2\theta.$