# Find derivatives: 1. y=(3x-1)^5 2. y=sqrt(x^2 +6x(sgrt x)) 3. y=ln^4 (8x-x^3) 4.f(x)=3^(4x^5+x-8) log2(6x+3)

sciencesolve | Teacher | (Level 3) Educator Emeritus

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1) You need to use power rule and chain rule to find derivative of `y=(3x-1)^5`  such that:

`(dy)/(dx)=5(3x-1)^4*(3x-1)'`

`(dy)/(dx)=3*5(3x-1)^4`

`(dy)/(dx)=15(3x-1)^4`

2) You need to use power rule, chain rule and product rule to find derivative of `y=sqrt(x^2 +6x(sqrt x))`  such that:

`(dy)/(dx) = ((x^2 +6x(sqrt x))')/(2sqrt(x^2 +6x(sqrt x)))`

`(dy)/(dx) = (2x +6(sqrt x) + 6x/(2sqrtx))/(2sqrt(x^2 +6x(sqrt x)))`

`(dy)/(dx) = (x +3(sqrt x) + 3x/(2sqrtx))/(sqrt(x^2 +6x(sqrt x)))`

3) You need to use power rule and chain rule to find derivative of `y=ln^4 (8x-x^3)`  such that:

`(dy)/(dx) = 4 ln^3 (8x-x^3)*((8x-x^3)')/(8x-x^3)`

`(dy)/(dx) = 4 ln^3 (8x-x^3)*(8-3x^2)/(8x-x^3)`

4) You need to change the base of logarithm `log_2 (6x+3)`  such that:

`log_ 2 (6x+3) = ln (6x+3)/ln 2`

You need to differentiate with respect to x such that:

`(dy)/(dx) =3^(4x^5+x-8)*ln 3*(4x^5+x-8)'(ln (6x+3)/ln 2)'`

`(dy)/(dx) =(ln 3/ln 2)*3^(4x^5+x-8)*(20x^4 + 1)*((6x+3)')/(6x+3)`

`(dy)/(dx) = (6*ln 3/ln 2)*3^(4x^5+x-8)*(20x^4 + 1)/(6x+3)`

`(dy)/(dx) = ((ln (3^6))/ln 2)*3^(4x^5+x-8)*(20x^4 + 1)/(6x+3)`