Derivative: find d/dx [(2x-3)^2(2x^2+1)^3]; simplify
- Thread starter smileykd
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skeeter
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let u = 2x-3 ... du/dx = 2
v = 2x^2 + 1 ... dv/dx = 4x
using the product rule and the chain rule ...
d/dx[u^2*v^3] = u^2*3v^2*(dv/dx) + v^3*2u*(du/dx)
factor out any common factors ...
d/dx[u^2*v^3] = u*v^2[3u*(dv/dx) + 2v*(du/dx)]
now, back substitute the expressions for u, v, and their derivatives ... simplify.
v = 2x^2 + 1 ... dv/dx = 4x
using the product rule and the chain rule ...
d/dx[u^2*v^3] = u^2*3v^2*(dv/dx) + v^3*2u*(du/dx)
factor out any common factors ...
d/dx[u^2*v^3] = u*v^2[3u*(dv/dx) + 2v*(du/dx)]
now, back substitute the expressions for u, v, and their derivatives ... simplify.