# Formulas for Higher-Order Derivatives

In general, to find n-th derivative of function y=f(x) we need to find all derivatives of previous orders. But sometimes it is possible to obtain expression for n-th derivative that depends on n and doesn't contain previous derivatives.

First of all it is easily to extend constant multiple, sum and difference rules.

Fact. For functions f(x) and g(x), and constant c we have that

• (cf)^((n))=c*f^((n));
• (f+-g)^((n))=f^((n))+-g^((n)).

Now consider power function y=x^a.

We have that y'=ax^(a-1), y''=a(a-1)x^(a-2).

In general, color(red)((x^a)^((n))=a(a-1)...(a-n+1)x^(a-n)).

Example 1 . Find y^((7)) if y=2x^5-5x^10.

We first use constant multiple and sum rule and then formula for power function.

y^((7))=(2x^5-5x^10)^((7))=2(x^5)^((7))-5(x^10)^((7))=

=2(5*4*3*2*1*0*(-1)*x^(5-7))-5(10*9*8*7*6*5*4*x^(10-7))=-3024000x^3.

For function y=a^x we have that y'=a^xln(a), y''=a^x(ln(a))^2 and in general color(green)((a^x)^((n))=a^x(ln(a))^n). In particular, color(blue)((e^x)^((n))=e^x).

For function y=sin(x) we have that y'=cos(x), y''=-sin(x), y'''=-cos(x), y^((4))=sin(x).

Since cos(x)=sin(x+pi/2) then color(107,32,35)((sin(x))^((n))=sin(x+n pi/2)).

Similarly, color(165,42,42)((cos(x))^((n))=cos(x+n pi/2)).

We saw that constant multiple and sum rules are easily extended for n-th derivative.

It is not a case with product rule.

So, let's see how product rule looks like in the case of higher-order derivatives.

Suppose that function f(x) and g(x) has finite derivatives up to n-th order.

We have that (fg)'=f'g+fg', (fg)''=(f'g+fg')'=(f'g)'+(fg')'=(f''g+f'g')+(f'g'+fg'')=f''g+2f'g'+fg''.

For third derivative we have (fg)'''=(f''g+2f'g'+fg'')'=(f''g)'+2(f'g')'+(fg'')'=

=(f'''g+f''g')+2(f''g'+f'g'')+(f'g''+fg''')=f'''g+3f''g+3f'g''+fg'''.

We see that these formulas are very similar to the expansion of binom of newton. Just instead of powers here are derivatives. If we denote f=f^((0)) and g=g^((0)) then we will obtain Leibniz Formula.

Leibniz Formula. (fg)^((n))=sum_(i=0)^n C_n^i f^((n-i))g^((i)).

We can extend this case when in product there are more than two functions, but this involves many formulas and calculations fairly complex.

Example 2. Find (x^2cos(x))^((50)).

Here g(x)=x^2 and f(x)=cos(x).

We have that, (f(x))^((k))=cos(x+k pi/2). Also g'(x)=2x, g''(x)=2, g'''(x)=g^((4))(x)=...=0.

This means that in Leibniz formula all summands except first three will be equal 0. Thus, (fg)^((50))=C_50^0 f^((50))g+C_50^1 f^((49))g'+C_50^2 f^((48))g''=

=1*cos(x+50 pi/2)*x^2+50*cos(x+49 pi/2)*2x+(50*49)/2 *cos(x+48 pi/2)*2=

=-x^2cos(x)-100xsin(x)+2450cos(x)=(2450-x^2)cos(x)-100xsin(x).

So, (x^2cos(x))^((50))=(2450-x^2)cos(x)-100xsin(x).