# 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

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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

(2x^3)'=2*3*x^(3-1)=6*x^2

(- 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

df(x)=d(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

2/x^2=-4x/x^4

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 | College Teacher | (Level 3) Valedictorian

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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:

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

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

neela | High School Teacher | (Level 3) Valedictorian

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

a)

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

b)

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

c)

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

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

d)

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)'

=2pi*(1/2)*(9.8/L)^(1/2)*(1/9.8)

=pi/sqrt(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)

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

=(1/9.8)pi/(L^(1/2)