Here we will cover the rules which we use for differentiating most types of function.
The derivative a function of the form $y=a$ (where $a$ is a constant) is zero: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=0\]
Note: This is intuitive as a constant function is a horizontal line which has a slope of zero.
To differentiate any function of the form: \[y=ax^n\] where $a$ and $n$ are constants, we take the power $n$, bring it in front of the function, and then reduce the power by $1$: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=n\times ax^{n-1}\]
Differentiate the function $y=x^4$.
\begin{align} \dfrac{\mathrm{d} y}{\mathrm{d} x}&=4\times x^{(4-1)}\\ &=4x^3 \end{align}
Differentiate the function $y=10x$.
\begin{align} \dfrac{\mathrm{d} y}{\mathrm{d} x}&=10\times1\times x^{1-1}\\ &=10x^0\\ &=10 \end{align}
Differentiate the function $y=2x^{~\frac{1}{4}~}$.
\begin{align} \dfrac{\mathrm{d} y}{\mathrm{d} x}&=2\times\left(\frac{1}{4}\right)\times x^{\left(\frac{1}{4}-1\right)}\\ &=\frac{1}{2}\times x^{\left(\frac{1}{4}-1\right)}\\ &=\frac{1}{2}\times x^{-\frac{3}{4}~}\\ &=\dfrac{1}{2 \sqrt[4]{x^3}~} \end{align} See Powers and Roots to remind yourself why $\frac{1}{2}x^{-\frac{3}{4}~}=\dfrac{1}{2 \sqrt[4]{x^3}~}$.
Differentiate the function $f(x)=3x^{-3}$.
\begin{align} f'(x)&=(-3)\times 3x^{-3-1}\\ &=-9x^{-4}\\ &=-\dfrac{9}{x^4}\\ \end{align}
To differentiate a sum (or difference) of terms, differentiate each term separately and add (or subtract) the derivatives.
Differentiate the function $y=10x^3+2x$.
First differentiate the first term in the sum, $10x^3$ using the power rule: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=30x^2\] Now differentiate the second term, $2x$ using the power rule again: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=2\] Adding these derivatives together give us the derivative of $y=10x^3+2x$: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=30x^2+2\]
Differentiate the function $g(x)=x^4-3x^{-3}$.
Let $m(x)=x^4$ and $n(x)=3x^{-3}$. We have already found the derivatives of these two functions. They are $m'(x)=4x^3$ and $n'(x)=-9x^-4$ respectively. The derivative of the function $g(x)=x^4+3x^{-3}$ is therefore the difference of these two derivatives: \begin{align} g'(x)&=3x^3-\left(-\dfrac{9}{x^4}\right)\\ &=3x^3+\dfrac{9}{x^4} \end{align}
Differentiate the function $y=4x^5+x^2-10x^{-2}-3$.
The derivative of the:
The derivative of $y=4x^5+x^2-10x^{-2}+3$ is therefore: \begin{align} \dfrac{\mathrm{d} y}{\mathrm{d} x}&=20x^4+2x-\left(-20x^{-3}\right)-0\\ &=20x^4+2x+20x^{-3} \end{align}
This rule is used to differentiate a function of another function, $y=f(g(x))$.To differentiate $y=f(g(x))$, let $u=g(x)$ so that we have $y$ as a function of $u$, $y=f(u)$. Then the chain rule says: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{\mathrm{d} y}{\mathrm{d} u}\times \dfrac{\mathrm{d} u}{\mathrm{d} x}\] Once you have worked this out, you replace $u$ by $g(x)$ and your answer is now in terms of $x$.
Differentiate the function $y=(3x+x^4)^2$.
The first step is to set $u=g(x)$, where $g(x)=3x+x^4$, and differentiate $u$ with respect to $x$: \[\dfrac{\mathrm{d} u}{\mathrm{d} x}=3+4x^3\] The next step is to differentiate $y$ with respect to $u$. Rewriting $y$ in terms of $u$ gives: \[y=u^2\] and so \[\dfrac{\mathrm{d} y}{\mathrm{d} u}=2u\] Next, using the chain rule, we have: \begin{align} \dfrac{\mathrm{d} y}{\mathrm{d} x}&=\dfrac{\mathrm{d} y}{\mathrm{d} u}\times \dfrac{\mathrm{d} u}{\mathrm{d} x}\\ &=2u\times (3+4x^3) \end{align} The final step is to substituting $u=3x+x^4$: \begin{align} \dfrac{\mathrm{d} y}{\mathrm{d} x}&=2u\times (3+4x^3)\\ &=2(3x+x^4)(3+4x^3) \end{align} After multiplying out the brackets and cancelling where possible, this simplifies to: \[2x(4x^6+15x^3+9)\]
Recall that the exponential function $f(x)= e^x$. The derivative of this function is the same as the function itself: \[\dfrac{\mathrm{d} f}{\mathrm{d} x}= e^x\] If the power to which $e$ is raised is a function of $x$, $g(x)$ say, we have $f(x)= e^{g(x)}$ and: \[\dfrac{\mathrm{d} f}{\mathrm{d} x}=g'(x) e^{g(x)}\] where $g'(x)$ denotes the derivative of the function $g(x)$.
Differentiate the function $f(h)= e^{h^2}$.
Here the power to which $ e$ is raised is a function so we have: \begin{align} \dfrac{\mathrm{d} f}{\mathrm{d} h}&=2h\times e^{h^2}\\ &=2h e^{h^2} \end{align}
We use the product rule to differentiate a function $y=uv$ which is a product of two functions of $x$, $u$ and $v$. The product rule says the derivative of $y$ is: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=u\dfrac{\mathrm{d} v}{\mathrm{d} x}+v\dfrac{\mathrm{d} u}{\mathrm{d} x}\]
In words this says that “the derivative of a product of two functions is the derivative of the first, times the second, plus the first times the derivative of the second.”
Differentiate the function $y=(1+2x^2)(3x+x^4)$.
Here $u=1+2x^2$ and $v=3x+x^4$. The derivative of $u$ with respect to $x$ is: \[\dfrac{\mathrm{d} u}{\mathrm{d} x}=4x\] and the derivative of $v$ with respect to $x$ is: \[\dfrac{\mathrm{d} v}{\mathrm{d} x}=3+4x^3\] Using the product rule, we have: \begin{align} \dfrac{\mathrm{d} y}{\mathrm{d} x}&=u\dfrac{\mathrm{d} v}{\mathrm{d} x}+v\dfrac{\mathrm{d} u}{\mathrm{d} x}\\ &=(1+2x^2)(3+4x^3)+(3x+x^4)4x \end{align} After multiplying out the brackets and cancelling where possible, this simplifies to: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=12x^5+4x^3+18x^2+3\]
Differentiate the function $y=(x^2+3x+6)(x^3+5x^2+2x+1)$.
Here $u=x^2+3x+6$ and $v=x^3-5x^2+2x-1$. The derivative of $u$ with respect to $x$ is: \[\dfrac{\mathrm{d} u}{\mathrm{d} x}=2x+3\] and the derivative of $v$ with respect to $x$ is: \[\dfrac{\mathrm{d} v}{\mathrm{d} x}=3x^2-5x+2\] Using the product rule, we have: \begin{align} \dfrac{\mathrm{d} y}{\mathrm{d} x}&=u\dfrac{\mathrm{d} v}{\mathrm{d} x}+v\dfrac{\mathrm{d} u}{\mathrm{d} x}\\ &=(x^2+3x+6)(3x^2-5x+2)+(x^3-5x^2+2x-1)(2x+3) \end{align} After multiplying out the brackets and cancelling where possible, this simplifies to: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=5x^4-3x^3-14x^2-18x+9\]
We use the quotient rule to differentiate a function $y=\dfrac{u}{v}$ which is a quotient of two functions of $x$, $u$ and $v$. The quotient rule says the derivative of $y$ is: \[\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{v\dfrac{\mathrm{d} u}{\mathrm{d} x}-u\dfrac{\mathrm{d} v}{\mathrm{d} x}}{v^2}\] In words, this says that the derivative of a quotient is “the derivative of the numerator times the denominator, minus the numerator times the derivative of the denominator, all divided by the denominator squared”
Differentiate the function $y=\dfrac{x^2+3}{x^3-2x+1}$
Here $u=x^2+3$ and $v=x^3-2x+1$. The derivative of $u$ with respect to $x$ is: \[\dfrac{\mathrm{d} u}{\mathrm{d} x}=2x\] and the derivative of $v$ with respect to $x$ is: \[\dfrac{\mathrm{d} v}{\mathrm{d} x}=3x^2-2\] Using the quotient rule, we have: \begin{align} \dfrac{\mathrm{d} y}{\mathrm{d} x}&=\dfrac{v\dfrac{\mathrm{d} u}{\mathrm{d} x}-u\dfrac{\mathrm{d} v}{\mathrm{d} x}~}{v^2}\\ &=\dfrac{(x^3-2x+1)2x-(x^2+3)(3x^2-2)}{(x^3-2x+1)^2} \end{align} After multiplying out the brackets in the numerator and cancelling where possible, this simplifies to \[\frac{dy}{dx}={-x^4-11x^2+2x+6}{(x^3-2x+1)^2}\]
Test yourself: Numbas test on differentiation
Test yourself: Numbas test on differentiation, including the chain, product and quotient rules