Calculus:Derivative of Polynomial Functions

Derivative of Polynomial Functions

=Newton Derivative of Polynomial Functions=

1. The sum rule $(f+g)'=f'+g'$
2. The Difference Rule $(f-g)'=f'-g'$
3. The Product Rule $(f*g)'=f*g'+g*f'$
4. The Quotient Rule $({f \over g})'={(gf'-fg') \over g^{2}}$

=Leibniz Derivative of Polynomial Functions=

1. The sum rule ${d(f+g) \over dx}={df \over dx}+{dg \over dx}$
2. The Difference Rule ${d(f-g) \over dx}={df \over dx}-{dg \over dx}$
3. The Product Rule ${d(fg) \over dx}'=f{dg \over dx}+g{df \over dx}$
4. The Quotient Rule ${d({f \over g}) \over dx}={g{df \over dx}-f{dg \over dx} \over g^{2}}$

Examples

Find the derivative

Example 1

Ex1:$f(x^{4}+2x^{2}+4x+2)$

$(f+g)'=f'+g'$

Using the sum rule we can divided in to different part $f'(x^{4}+2x^{2}+4x+2)=(x^{4})+2(x^{2})+4(x)+(2)$

so we will started to work on different part by using power rule.

$f'(x)=(x^{4})+2(x^{2})+4(x)+(2)$

$(4x^{3})+2(2x)+4$

$4x^{3}+4x+4$

Example 2

Ex2:$f(x)=x^{4}*x^{3}$

The Product Rule $(f*g)'=f*g'+g*f'$

Using the Product Rule we can divided in to different part $f'(x^{4}*x^{3})=x^{4}*(x^{3})'+(x^{4})'*x^{3})$

$x^{4}*3x^{2}+4x^{3}*x^{3}$

$3x^{6}+4x^{6}$

$3x^{6}+4x^{6}$

$7x^{6}$

Example 3

Ex3:$z(v)={{4v^{4}} \over {v^{3}+5v}}$

Now we can understand v as x the idea will be the same.

By using the quotient rule $({f \over g})'={(gf'-fg') \over g^{2}}$

we can under stand it as

f(v)=4v^4 g(v)=v^3 + 5v

so we will get

Failed to parse (syntax error): {\displaystyle ]z'(v) =(v^3 + 5v)(4v^4)' - (4v^4)(v^3 + 5v)'\over (v^3 + 5v)^2 } }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ]z'(v) =(v^3 + 5v)(16v^3) - (4v^4)(v^3 + 5v)'\over (v^3 + 5v)^2 } }