# Differentiation of Polynomials

Polynomials, as the term implies, contain multiple terms, added or subtracted together. In order to differentiate the polynomial, each term must be differentiated separately. Here we will look at differentiating polynomials of the form,

Polynomials, as the term implies, contain multiple terms, added or subtracted together. In order to differentiate the polynomial, each term must be differentiated separately. Here we will look at differentiating polynomials of the form,

(y=a_{0}+a_{1}x^{1} +a_{2}x^{2} +….+ a_{n}x^{n})

Definition

If (y=x^{n}) then (frac{dy}{dx}=nx^{n-1})

An example
a. Differentiate, (y=x^{7}).
b. Find the gradient to the curve (y=x^{7}) when (x=2).
c. Find the equation of the tangent to the curve when (x=2).
(a) Using the formula above we see that (n=7) and therefore

[begin{eqnarray*} frac{dy}{dx}=7x^{7-1}=7x^{6} end{eqnarray*}]

(b) From part (a) we know that (frac{dy}{dx}=7x^{6}). Therefore when (x=2) the gradient is (frac{dy}{dx}=7times 2^{6}=448).
(c) When (x=2), (y=x^{7} =2^{7} =128). In part (b) we found that the gradient of the tangent is 448.
The equation of a straight line is (y=mx+c) where (m) is the gradient and (c) is the y-intercept. Thus (y=448x+c). Using our values of (x) and (y),

[begin{eqnarray*} 128=448times 2 +c 128=896+c c=-768 end{eqnarray*}]

Therefore the equation of the tangent to the curve is (y=448x-768).
Definition
If (y=cx^{n}) then (frac{dy}{dx}=cnx^{n-1})
An example
a. Differentiate (y=4x^{5})
b. Find the gradient to the curve (y=4x^{5}) when (x=-2)
c. Find the equation of the tangent to the curve when (x=-2).
(a) Using the formula above we see that (c=4) and (n=5). Therefore

[begin{eqnarray*} frac{dy}{dx}=4times 5times x^{5-1}=20x^{4} end{eqnarray*}]

(b) From part (a) we found (frac{dy}{dx}=20x^{4}). Therefore when (x=-2) we calculate that the gradient to the curve at (x=-2) is (frac{dy}{dx}=20times (-2)^{4}=320).
(c) When (x=-2), (y=4x^{5}=4times (-2)^{5}=-128). In part (b) we found that the gradient of the tangent is 320.
The equation of a straight line is (y=mx+c) where (m) is the gradient and (c) is the y-intercept. Thus (y=320x+c). Using our values of (x) and (y),

[begin{eqnarray*} -128=320times -2 +c -128=-640+c c=512 end{eqnarray*}]

Therefore the equation of the tangent to the curve is (y=320x+512).
Definition
If (y=f(x)+g(x)), then (frac{dy}{dx}=frac{df(x)}{dx}+frac{dg(x)}{dx})
An example
Differentiate (y=3x^{2}-4x +7 -3x^{-2}).
Using the formula above we can see that we can differentiate each part separately so that
(y=f(x)+g(x)+h(x)+j(x)) where (f(x)=3x^{2}), (g(x)=-4x) (h(x)=7) and (j(x)=-3x^{-2}).
We first find (frac{df(x)}{dx}).

[begin{equation*} frac{df(x)}{dx}=3times 2 times x^{2-1}=6times x^{1}=6x end{equation*}]

Next we find

[begin{equation*} frac{dg(x)}{dx}=-4times 1times x^{1-1}=-4x^{0}=-4 end{equation*}]

Then

[begin{equation*} frac{dh(x)}{dx}=0 end{equation*}]

because from first principles we find that if (y=c) then (frac{dy}{dx}=0).
Finally

[begin{equation*} frac{dj(x)}{dx}=-3times -2 times x^{-2-1}=6x^{-3} end{equation*}]

Putting all of this together we find

[begin{eqnarray*} frac{dy}{dx}=frac{df(x)}{dx}+frac{dg(x)}{dx}+frac{dh(x)}{dx}+frac{dj(x)}{dx} =6x-4+0+6x^{-3}=6x-4+6x^{-3} end{eqnarray*}]

## Over to you:

Find the gradient of the curve (y=-4x^{-2}) at the point where (y=-1) and (x>0.)

What common mistakes do you think students make when answering this question?