Let,

`y = 3x^2(2x-5)^4`

To differentiate with respect to x, apply the power formula of derivatives which is `d/(dx)(f(x)*g(x)) = f(x) g'(x) + g(x) f'(x)` .

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

To get `((2x-5)^4)'` and `(3x^2)'` , use the power formula of derivatives which is `d/(dx) (u^n) = n*u^(n-1)*u'` .

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Let,

`y = 3x^2(2x-5)^4`

To differentiate with respect to x, apply the power formula of derivatives which is `d/(dx)(f(x)*g(x)) = f(x) g'(x) + g(x) f'(x)` .

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

To get `((2x-5)^4)'` and `(3x^2)'` , use the power formula of derivatives which is `d/(dx) (u^n) = n*u^(n-1)*u'` .

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

`(dy)/(dx) = 12x^2(2x-5)^3(2x-5)' + 6x(2x-5)^4 x'`

Note that:

`(2x-5)' = d/(dx)(2x-5)= d/(dx) (2x)- d/(dx)5 = 2 - 0 =2 `

`x' = d/(dx) (x) = 1`

So,

`(dy)/(dx)=12x^2(2x-5)^3*2 + 6x(2x-5)^4*1`

`(dy)/(dx)=24x^2(2x-5)^3 + 6x(2x-5)^4`

**Hence, `(dy)/(dx)=24x^2(2x-5)^3+6x(2x-5)^4` .**