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Hi, I need help solving this equation using the 'Product Rule', thanks

Hi, I need help solving this equation using the 'Product Rule', thanks-example-1

1 Answer

6 votes

Given:


y=2x(x+3)^2

Where:

Gradient = 14

Let's find the exact values of x for which the gradient of the tangent line to the curve is 14.

Let's first simplify the equation:


y=2x((x+3)(x+3))

Expand using the FOIL method:


\begin{gathered} y=2x(x(x+3)+3(x+3)) \\ \text{ } \\ \text{ Apply distributive property:} \\ y=2x((x(x)+x(3)+3(x)+3(3)) \\ \\ y=2x(x^2+3x+3x+9) \\ \\ y=2x(x^2+6x+9) \\ \\ y=2x(x)+2x(6x)+2x(9) \\ \\ y=2x^3+12x^2+18x \end{gathered}

Now, let's find the first derivative of the equation:


\begin{gathered} (d)/(dx)(y)=(d)/(dx)(2x^3+12x^2+18x) \\ \\ y^(\prime)=6x^2+24x+18 \end{gathered}

Set the derivative to 14 and solve for x.

We have:


y^(\prime)^(\prime)=12x+24

Now, set the second derivative to 14:


12x+24=14

Solve for x:

Subtract 24 from both sides:


\begin{gathered} 12x+24-24=14-24 \\ \\ 12x=-10 \\ \\ x=-(10)/(12) \\ \\ x=-(5)/(6) \end{gathered}

Therefore, the exact value of x for which the gradient of the tangent to the curve is given is:


-(5)/(6)

ANSWER:


x=-(5)/(6)

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