find y' for y=(2x+1)^10 20(2x+1)^10 10(2x+1)^9 20(2x+1)^9 20(2x-1)^9

To find the derivative \( y' \) of the function \( y = (2x+1)^{10} \), we will use the chain rule. The chain rule states that if you have a composite function \( y = f(g(x)) \), then the derivative \( y' \) is given by:<br /><br />\[ y' = f'(g(x)) \cdot g'(x) \]<br /><br />In this case, let \( u = 2x + 1 \). Then \( y = u^{10} \).<br /><br />First, we find the derivative of \( y \) with respect to \( u \):<br /><br />\[ \frac{dy}{du} = 10u^9 \]<br /><br />Next, we find the derivative of \( u \) with respect to \( x \):<br /><br />\[ \frac{du}{dx} = 2 \]<br /><br />Now, applying the chain rule:<br /><br />\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 10u^9 \cdot 2 \]<br /><br />Substitute \( u = 2x + 1 \) back into the expression:<br /><br />\[ \frac{dy}{dx} = 10(2x+1)^9 \cdot 2 = 20(2x+1)^9 \]<br /><br />Therefore, the correct answer is:<br /><br />\[ 20(2x+1)^9 \]