Use the quotient rule. u = (2x - 3)^2 and v = (x^3 - 7)^3. To differentiate u and v, either expand and differentiate in the usual way, or use the chain rule.
Just in case you're interested (like many people) in avoiding the quotient rule, which can feel like overkill, and always doubles the power of the denominator, sometimes unnecessarily as here... then, how about...
... where
... is the product rule, straight lines differentiating downwards with respect to x. Then you don't need u and v. You still need the chain rule, but you can zoom in on this, for example in the right hand fork...
... where...
... is the chain rule. Straight continuous lines still differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
However, you could zoom in on both forks...
... or neither...
... according to taste.
Another way of avoiding the quotient rule is logarithmic differentiation - taking logs of both sides of y = the function. Then solve for dy/dx in the bottom row of...
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\(\displaystyle f(x)=(2x-3)^2/(x^3- 7)^3 \)
\(\displaystyle f' (x)=((x^3-7)^3 [2(2x-3)(2) ]-[3(x^3-7)^2 (3x^2 )])/[(x^3-7)^3 ]^2 \)
\(\displaystyle f' (x)=((x^3-7)^3 [4(2x-3)]-[9x^2 (x^3-7)^2](2x-3)^2)/(x^3-7)^6 \)
We need to simplify:
\(\displaystyle f' (x)=((x^3-7)[4(2x-3) ]-(9x^2)(2x-3)^2)/(x^3-7)^4 \)
We need to simplify Further:
\(\displaystyle f' (x)=(2x-3)[4(x^3-7)-9x^2 (2x-3) ]/(x^3-7)^4 \)
thats my final answer ^^...I am supposed to represent the derivative in its simplified form..Is my working correct?