How do you find the derivative of #y=(2x^3+4)(x^2-3x+1)#?
1 Answer
Jan 29, 2017
Explanation:
differentiate using the
#color(blue)"product rule"#
#"Given "y=f(x).g(x)" then"#
#color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=f(x)g'(x)+g(x)f'(x))color(white)(2/2)|)))larr" product rule"#
#"here "f(x)=2x^3+4rArrf'(x)=6x^2#
#"and "g(x)=x^2-3x+1rArrg'(x)=2x-3#
#rArrdy/dx=(2x^3+4)(2x-3)+(x^2-3x+1).6x^2# distributing brackets.
#=4x^4-6x^3+8x-12+6x^4-18x^3+6x^2#
#rArrdy/dx=10x^4-24x^3+6x^2+8x-12#