Question

The differentiation of function f(x) = (3x + 2)^3/2 w.rt. x is​

Answer 1

[Note : refer the attachment for the solution of the problem ]

$\boxed{\textbf{\large{step bystep explanation}}}$

◾The given functions is,

y = (3x + 2)^3/2

differentiation wrt x

dy /dx = d/dx ( (3x + 2 ) ^(3/2))

◾here first differentiate the term ( 3x + 2 ) ^(3 /2 ) wrt x, we know derivative of standard function

y =[f(x)]^n

dy /dx =nf[( x)]^( n - 1)x f'(X) ,

so from this we can differentiate ( 3x + 2 )^( 3/ 2) as a

= 3 / 2 ( 3x + 2 )^( (3/2 ) - (1)) ✖ (d/dx( 3x + 2))

◾After this then differentiate ( 3x + 2) As we know the rule ofdifferentiation,

If y = ( u + v) , then

dy/dx = d/dx ( u + v)

=(du /dx)+(dv/dx)

so from above first differentiate 3x and then differentiate 2 in the function ( 3x + 2 )

when we differentiate a standard function kx then derivative is K, so from this derivative of 3x is 3 and derivative of constant is always zero therefor derivative of 2 is zero .

d/dx( 3x + 2 ) = ( 3 + 0 ) = 3

= 3/2 ( 3x + 2 )^( 1/2 ) ✖ ( 3 )

◾So After this do the simple calculations for final result .

= 9/2 (( 3x + 2 )^( 1/ 2))

= [ 9 ( 3x + 2 )^( 1/2 ) ]/ 2

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here's the derivatives of some standard composite functions

1] y =[ f(x) ]^n

➡ dy / dx = n[f(x)]^n ✖ f'(x )

2] y = √ ( f (x ) )

➡dy / dx =( 1 / √[ f(x)]^2 ) ✖ f'(x )

3] y = sin f (x)

➡dy / dx = cos f (x) ✖ f' (x)

4] y = cos f (x )

➡ dy /dx = -sin f (x) ✖ f'(x )

5] y = k

➡ dy / dx = 0 [ Note : it is not a composite function ]

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Answer 2

3/2(3x+2)^(3/2-1)* 3
= 3/2(3x+2)^1/2*3