Find dy/dx such that y=(e^x)(3x+1)^2.

We will solve this question with the knowledge that dy/dx = u.(dv/dx) + v.(du/dx), where y=u.v We have y=e^x(3x+1)^2. First, we want to find u & v. By splitting the function, we have that u=e^x and v=(3x+1)^2. Then, we want to find du/dx and dv/dx. In other words, differentiate u & v wrt x. Since e^x differentiated wrt x is just e^x, du/dx=e^x. To differentiate (3x+1)^2, we must differentiate the function inside and outside of the brackets. So, we have 2(3x+1) multiplied by 3. This gives us 6(3x+1). Hence dv/dx=6(3x+1). Using our original equation for the derivative, dy/dx = e^x.(6(3x+1)) + (3x+1)^2.(e^x) Therefore, dy/dx = 6e^x(3x+1) +e^x(3x+1)^2. Finally, we must not forget to add a constant to our answer. Hence, our final answer is dy/dx = 6e^x(3x+1) +e^x(3x+1)^2 + c.

SB

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