(2x+3)^3(x-4) How do I differentiate this?
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Jan 27, 2005 #2 courtrigrad 1,233 1 Use the product rule: [itex] vdu + udv [/itex]. For the first function use the power rule: [itex] x^n = nx^{n-1} \frac{du}{dx}=(3(2x+3)^2(2) [/itex] Last edited: Jan 27, 2005
Use the product rule: [itex] vdu + udv [/itex]. For the first function use the power rule: [itex] x^n = nx^{n-1} \frac{du}{dx}=(3(2x+3)^2(2) [/itex]
Jan 27, 2005 #5 arildno Science Advisor Homework Helper Gold Member Dearly Missed 9,948 130 No! You must be more careful! Define: [tex]f(x)=(2x+3)^{3},g(x)=(x-4)[/tex] Then, [tex]y(x)=f(x)g(x),y'(x)=f'(x)g(x)+f(x)g'(x)[/tex] You need therefore to calculate f'(x) and g'(x)
No! You must be more careful! Define: [tex]f(x)=(2x+3)^{3},g(x)=(x-4)[/tex] Then, [tex]y(x)=f(x)g(x),y'(x)=f'(x)g(x)+f(x)g'(x)[/tex] You need therefore to calculate f'(x) and g'(x)