Question

If $f(x) = (2x-3\pi)^{25} + \frac{4}{3}x + \cos x$ and $g(x)$ is inverse of $f(x)$ then find $\frac{d}{dx}(g(x))$ at $x = 2\pi$.

Answer 1

$\frac{3}{7}$

Answer 2

Let $f(x) = (2x-3\pi)^{25} + \frac{4}{3}x + \cos x$. We are given that $g(x)$ is the inverse of $f(x)$. We need to find $\frac{d}{dx}(g(x))$ at $x = 2\pi$, which is $g'(2\pi)$.

The formula for the derivative of an inverse function states that if $g$ is the inverse of $f$, then $g'(y) = \frac{1}{f'(x)}$ where $y = f(x)$. In this problem, we want to find $g'(2\pi)$. So, we need to find a value $x_0$ such that $f(x_0) = 2\pi$.

Let's substitute $x = \frac{3\pi}{2}$ into $f(x)$: $f\left(\frac{3\pi}{2}\right) = \left(2\left(\frac{3\pi}{2}\right) - 3\pi\right)^{25} + \frac{4}{3}\left(\frac{3\pi}{2}\right) + \cos\left(\frac{3\pi}{2}\right)$ $f\left(\frac{3\pi}{2}\right) = (3\pi - 3\pi)^{25} + \frac{12\pi}{6} + 0$ $f\left(\frac{3\pi}{2}\right) = (0)^{25} + 2\pi + 0$ $f\left(\frac{3\pi}{2}\right) = 2\pi$ So, we have found that $x_0 = \frac{3\pi}{2}$ such that $f(x_0) = 2\pi$.

Now, we need to find the derivative of $f(x)$, denoted as $f'(x)$. $f'(x) = \frac{d}{dx}\left((2x-3\pi)^{25} + \frac{4}{3}x + \cos x\right)$ Using the power rule and chain rule: $f'(x) = 25(2x-3\pi)^{24} \cdot \frac{d}{dx}(2x-3\pi) + \frac{d}{dx}\left(\frac{4}{3}x\right) + \frac{d}{dx}(\cos x)$ $f'(x) = 25(2x-3\pi)^{24} \cdot 2 + \frac{4}{3} - \sin x$ $f'(x) = 50(2x-3\pi)^{24} + \frac{4}{3} - \sin x$

Next, we evaluate $f'(x)$ at $x_0 = \frac{3\pi}{2}$: $f'\left(\frac{3\pi}{2}\right) = 50\left(2\left(\frac{3\pi}{2}\right)-3\pi\right)^{24} + \frac{4}{3} - \sin\left(\frac{3\pi}{2}\right)$ $f'\left(\frac{3\pi}{2}\right) = 50(3\pi-3\pi)^{24} + \frac{4}{3} - (-1)$ $f'\left(\frac{3\pi}{2}\right) = 50(0)^{24} + \frac{4}{3} + 1$ $f'\left(\frac{3\pi}{2}\right) = 0 + \frac{4}{3} + \frac{3}{3}$ $f'\left(\frac{3\pi}{2}\right) = \frac{7}{3}$

Finally, we use the inverse function derivative formula: $g'(2\pi) = \frac{1}{f'(x_0)} = \frac{1}{f'(3\pi/2)}$ $g'(2\pi) = \frac{1}{7/3}$ $g'(2\pi) = \frac{3}{7}$