Category Archives: Quotient Rule

February 4, 2025 · 9:17 am

Differentiating the Tangent Function

Remember tan(x)=\frac{sin(x)}{cos(x)}.

I use the quotient rule to differentiate f(x)=tan(x).

(1)   \begin{equation*}\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\end{equation*}

If h(x)=tan(x)=\frac{sin(x)}{cos(x)} then from equation 1

(2)   \begin{equation*}h'(x)=\frac{cos(x)cos(x)-(-sin(x)sin(x))}{[cos(x)]^2}\end{equation*}

(3)   \begin{equation*}h'(x)=\frac{cos^2(x)+sin^2(x)}{cos^2(x)}\end{equation*}

Remember the Pythagorean identity

(4)   \begin{equation*}sin^2(x)+cos^2(x)=1\end{equation*}

Hence

    \begin{equation*}h'(x)=\frac{1}{cos^2(x)}=sec^2(x)\end{equation}

(5)   \begin{equation*}\frac{d}{dx}tan(x)=sec^2(x)\end{equation*}

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Filed under Calculus, Differentiation, Differentiation, Identities, Quotient Rule, Trigonometry, Year 12 Mathematical Methods

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January 14, 2025 · 7:52 am

Deriving the Quotient Rule for Differentiation

Like we did for the product rule, we are going to derive the differentiating rule for functions in the form y=\frac{f(x)}{g(x)}.

Something like, y=\frac{x^2+3x+2}{x^3-1}

Remember the first principals limit

\lim_{\limits h \to 0}\frac{f(x+h)-f(x)}{h}

If y=\frac{f(x)}{g(x)}, then

y'=\lim_{\limits h \to 0}\frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}

Find a common denominator for the numerator (i.e. g(x+h)g(x))

y'=\lim_{\limits h \to 0}\frac{\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}}{h}

To make things a bit easier I am going to multiply by \frac{1}{h} rather than having h as the denominator

y'=\lim_{\limits h \to 0}\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)} \times \frac{1}{h}

Now I am going to add and subtract f(x)g(x)

y'=\lim_{\limits h \to 0}\frac{f(x+h)g(x)-f(x)(g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)} \times \frac{1}{h}

Factorise

y'=\lim_{\limits h \to 0}\frac{g(x)(f(x+h)-f(x))+f(x)(g(x)-g(x+h))}{g(x+h)g(x)} \times \frac{1}{h}

Change the sign in the middle

y'=\lim_{\limits h \to 0}\frac{g(x)(f(x+h)-f(x))-f(x)(g(x+h)-g(x))}{g(x+h)g(x)} \times \frac{1}{h}

Separate the limits

y'=g(x)\lim_{\limits h \to 0}\frac{\frac{f(x+h)-f(x)}{h}}{g(x+h)g(x)}-f(x)\lim_{\limits h \to 0}\frac{\frac{g(x+h)-g(x)}{h}}{g(x+h)g(x)}

which simplifies to

y'=g(x)\frac{f'(x)}{g(x)g(x)}-f(x)\frac{g'(x)}{g(x)g(x)}

y'=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}

In words

The derivative of the top times the bottom take the derivative of the bottom times the top all over the bottom squared

Example

y=\frac{x^2+3x+2}{x^3-1}

y'=\frac{(2x+3)(x^3-1)-3x^2(x^2+3x+2)}{(x^3-1)^2}

y'\frac{2x^4-2x+3x^3-3-3x^4-9x^3-6x^2}{(x^3-1)^2}

y'=\frac{-x^4-6x^3-6x^2-2x-3}{(x^3-1)^2}

Exam questions usually specify no simplifying.

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Filed under Calculus, Differentiation, Quotient Rule, Year 12 Mathematical Methods