Using the rules of differentiation, find the derivative of each of the following functions    a) y=2x^3 - 7x +5 b) f(x) = 2/x^2 c) f(x) = 4sqrt x d) g(L) = 2pi sqrt L/9.8

4 Answers

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Using the rule of differentiation, for the first function, where

a) y=2x^3 - 7x +5, we could write dy=(2x^3 - 7x +5)dx (you could read this in this way, the derivative of the function from the left side of the equal is made  having y as unknown and the derivative of the function from the right side of the equal, 2x^3 - 7x +5, is made having x as unknown. We could write as (y)'=(2x^3 - 7x +5)')

dy=(2x^3 - 7x +5)dx=(2x^3)dx+( - 7x)dx +(5)dx

(2x^3) is a power function and it's derivative is


(- 7x) is a linear function, where it's derivative is

( - 7x)'=-7*1*x^( 1-1)=-7

+5 is a constant functio, where it's derivative is 0

So,  dy=(6*x^2-7)dx

b) f(x) = 2/x^2


 2/x^2 could be seen as a ratio and the derivative of a ratio is: the numerator derivative multiplied with the denominator minus the denominator derivative multiplied with the numerator, all these divided to the denominator square raised.

So  2/x^2=[ (2)'*(x^2)-(2)*(x^2)']/(x^2)^2


Simplifying the unknown x both the numerator and denominator, we'll obtain:

 d( 2/x^2)=(-4/x^3)dx

c)f(x) = 4sqrt x

df=d(4sqrt x)

sqrt x= 1/2sqrt x, sqrt x could be seen as a power function, too; sqrt x= (x)^1/2, so, [(x)^1/2]'=1/2*(x)^(1/2-1)=

=1/2*(x)^(-1/2)=(1/2)/(x)^1/2=(1/2)/sqrt x=1/2sqrt x

d(4sqrt x)=(4sqrt x)'=4*(sqrtx)'=4*(1/2sqrt x)=2/sqrt x.

it's improper to have a square root at denominator, so, we'll amplify with the same value of the square root, in order to have a denominator without square root.

d(4sqrt x)=(2*sqrt x/x)


d)g(L) = 2pi sqrt L/9.8

d(g(L))= d(2pi sqrt L/9.8)

If you are looking at the function above, the only difference is that the unknown is L and not x, the unknown we're used to. You can re-write the function g(L) in a more  intelligible way, more accurate, so g(L)= (2pi/sqrt 9.8)*sqrtL.

All you have to do is to consider sqrtL as  sqrt x=1/2sqrt x, so sqrt L=1/2sqrtL

d(g(L))=[(2pi/sqrt 9.8)*sqrtL]'=(2pi/sqrt 9.8)*1/2sqrtL=

d(g(L))=  (pi *sqrtL/L*sqrt 9.8)dx


krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

Before we start finding derivatives of the expressions given in the question, let us revise to rules of finding derivatives that we will need.

  1. Derivative of a function a*x^n is: a*[nx^(n-1)].
  2. Derivative of a constant is zero.

Now we can find the derivative of the four expressions given in the questions.

a) y=2x^3 - 7x +5

Derivative of 2x^3 - 7x + 5 is:

2*3x^2 - 7 = 6x^2 - 7

b) f(x) = 2/x^2

Derivative of 2*x^(-2) is:

2*(-2)*[x^(-3)] = -4/(x^3)

c) f(x) = 4sqrt x

Derivative of 4x^(1/2) is:

4*(1/2)*x^(-1/2) = 2/[x^1/2)]

d) g(L) = 2pi sqrt L/9.8

Derivative of (L/^1/2)/[9.8^(1/2)] is:


= {pi*/[9.8^(1/2)]}/(L^1/2) = pi/[(9.8*L)^(1/2)]

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on


If y=x^n, dy dx =nx^(n-1). Using this rule, we solve :


given y = 2x^3-7x+5.

dy/dx= d/dx(2x^3-7x+5)=d/dx(2x^3-7x^1+5*x^0)

2*3x^(3-1)-7*1*x^(1-1)+5*0*x^(0-1) = 6x^2-7


f(x) = 2/x^2=2*x^(-2)

df(x)/dx=f'(x)= 2*(-2-1)*x^(-2-1)= -6*x^(-3) = -6/x^3


f(x) = 4sqrtx =4*x^(1/2)

f'(x)=4*(1/2)*x^(1/2 - 1) = 2/x^(1/2)


Rule: f'(u(x)) = f'(u)*u'(x)

g(L) = 2pi sqrt(L/9.8)

g(L) = 2pi* sqrt(u(L)) , where u(L) = ( L/9.8)

g'(L) = 2pi* (1/2)(L/9.8)^(1/2 -1)*(L/9.8)'




I made a correction in (d) with a bracket. If bracket is not there, then

g(L)=2pi*sqrtL/9.8 = (2/9.8)pi* sqrtL = (2/9.8)pi*L^0.5

g'(L) = (2/9.8)pi*(0.5)*L^(0.5-1)