Given that y = 3x(^2) + 6x(^1/3) + (2x(^3) - 7)/(3(sqrt(x))) when x > 0 find dy/dx

Firstly, the (2x(^3) - 7)/(3(sqrt(x))) can be split into (2x(^3))/(3(sqrt(x)) and -7/(3(sqrt(x)). These can then be simplified to (2/3)x(^5/2) and -(7/3)x(^-1/2) respectively. This then gives the equation y = 3x(^2) + 6x(^1/3) + (2/3)x(^5/2) - (7/3)x(^-1/2).

By multiplying the coefficients of x by the power of x and then taking 1 from the power it is found that dy/dx = 6x + 2x(^-2/3) + (5/3)x(^3/2) + (7/6)x(^-3/2).

SH

Related Maths A Level answers

All answers ▸

I struggle to simplify the following equation: (see answer)


Find the minimum and maximum points of the graph y = x^3 - 4x^2 + 4x +3 in the range 0<=x <= 5.


Given y = 2x(x^2 – 1)^5, show that dy/dx = g(x)(x^2 – 1)^4 where g(x) is a function to be determined.


Differentiate the following equation with respect to x; sinx + 3x^2 - 2.