How do I find the derivative of the function $f(x)= (2x+1)^2$? I've tried doing this problem and am not fully sure that I am correct. I found the derivative to be $f'(x) = 8x+4$. Is that correct?
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$\begingroup$Yes, the derivative is correct.
To insert the details, note that $f(x) = (h \circ g)(x)$ with $$g(x) = 2x + 1$$ and $$h(u) = u^2$$ Then the chain rule simply says that
$$f'(x) = h'(g(x)) g'(x)$$
Now $h'(u) = 2u$ and so $h'(g(x)) = 2 g(x) = 2(2x + 1)$. On the other hand, $g'(x) = 2$; putting it together, we get
$$f'(x) = 2(2x + 1) 2 = 8x + 4$$
as you've stated.
$\endgroup$ 0 $\begingroup$Remember that you apply the chain rule when you have an expression $f(g(x))$. In this case, $g(x) = (2x+1)$, and your $f(a) = a^2$. The chain rule says that $f(g(x))' = f'(g(x))\cdot g'(x)$. Work out the derivatives: $g'(x) = 2$, $f'(x) = 2a$. Substitute $g'(x)$ for $a$ and put it all together: $2(2x+1)\cdot2 = 4(2x+1) = 8x+4$.
Also, this question is essentially a duplicate. Please refrain from asking questions that have already been asked before.
$\endgroup$ $\begingroup$The Chain Rule: If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$, then the composite function $F=f\circ g$ defined by $F(x)=f(g(x))$ is differentiable at x and $F'$ is given by the product $$F'(x)=f'(g(x))\cdot g'(x).$$If $f(x)=(2x+1)^2$, then $f'(x)=2(2x+1)2=4(2x+1)=8x+4$.
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