MATH1011 - Lecture 4

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MATH1011 - Lecture 3
Skeleton Slides

Differentiation Recap

• Introduction to Differential Calculus
• Applications of Differential Calculus
• Calculus Y12

Differentiable Functions are Continuous

• If $f$ is differentiable at some $c\in I$, then $f$ is continuous at $c$
  • The inverse of this statement is not true
  • i.e. not every continuous function is differentiable
• ($I$ means interval)

We say a function is smooth on some domain $D$ is the derivatives of all order exist on $D$

Mean Value Theorem

• For an interval $[a,b]$ where $(a,b)$ is differentiable, there exists a point where its derivative is parallel to the slope formed between both ‘corners’ of the interval
  • The slope from $(a,f(a))$ and $(b,(b))$ forms a diagonal line
  • A point within than interval, $c$, has the same gradient
• Watch lecture recording for full explanation

Read: Inverse Functions

Derivative of Vector-Valued Functions

• All rules that apply to normal differentiation also apply to vector-valued function differentiation (e.g. chain rule)
• However, there is two types of product rule:
  • Scalar Product Rule $[f(t)u(t){]}^{\prime }=f^{\prime }(t)u(t)+f(t)u^{\prime }(t)$
  • Dot Product Rule $[u(t)\cdot v(t){]}^{\prime }=u^{\prime }(t)\cdot v(t)+u(t)\cdot v^{\prime }(t)$

Derivative of Inverse Trigonometric Functions

• $\frac{d}{dx}({\sin }^{-1}x)=\frac{1}{\sqrt{1-x^2}}$
• $\frac{d}{dx}({\cos }^{-1}x)=-\frac{1}{\sqrt{1-x^2}}$
• $\frac{d}{dx}({\tan }^{-1}x)=\frac{1}{1+x^2}$
• Tutorial and Proof

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Annotated Slides