Differentiating Powers of x (DP IB Applications & Interpretation (AI): SL): Revision Note

Differentiating Powers of x

What is differentiation?

• Differentiation is the process of finding an expression for the derivative (gradient function) of a function from an expression for the function

How do I differentiate powers of x?

• Powers of $x$ are differentiated according to the following formula:

  • If$f\left(x\right)=x^n$ then$f^{\text{'}}\left(x\right)=n{x}^{n-1}$ where $n\in \mathrm{\mathbb{Z}}$

  • This is given in the formula booklet

• If the power of $x$ term is multiplied by a constant then the derivative is also multiplied by that constant

  • If$f\left(x\right)=a{x}^n$ then$f^{\text{'}}\left(x\right)=an{x}^{n-1}$ where $n\in \mathrm{\mathbb{Z}}$ and $a$ is a constant

• The alternative notation (to$f^{\text{'}}\left(x\right)$) is to use $\frac{\mathrm{d}y}{\mathrm{d}x}$

  • If $y=a{x}^n$ then $\frac{\mathrm{d}y}{\mathrm{d}x}=an{x}^{n-1}$

    • e.g.  If $y=-4x^5$ then $\frac{\mathrm{d}y}{\mathrm{d}x}=-4\times 5\times x^{5-1}=-20x^4$

• Don't forget these two special cases:

  • If$f\left(x\right)=ax$ then$f^{\text{'}}\left(x\right)=a$

    • e.g.  If $y=6x$ then $\frac{\mathrm{d}y}{\mathrm{d}x}=6$

  • If$f\left(x\right)=a$ then$f^{\text{'}}\left(x\right)=0$

    • e.g.  If $y=5$ then $\frac{\mathrm{d}y}{\mathrm{d}x}=0$

• Functions involving fractions with denominators in terms of $x$ will need to be rewritten as negative powers of $x$ first

  • e.g.  If$f\left(x\right)=\frac{4}{x}$ then rewrite as$f\left(x\right)=4x^{-1}$ before differentiating

How do I differentiate sums and differences of powers of x?

•  For an expression that is a sum or difference of powers of $x$, just differentiate term by term

  • e.g.  If$f\left(x\right)=5x^4+2x^3-3x+4$ then
    $f^{\text{'}}\left(x\right)=5\times 4x^{4-1}+2\times 3x^{3-1}-3+0$
    $f^{\text{'}}\left(x\right)=20x^3+6x^2-3$

• Products and quotients cannot be differentiated in this way so would need to be expanded/simplified first

  • e.g.  If$f\left(x\right)=(2x-3)(x^2-4)$ then expand to$f\left(x\right)=2x^3-3x^2-8x+12$ which is a sum/difference of powers of $x$ and can be differentiated