Question
\frac{d}{dx}\left(3+cos^2\left(z\right)\right)^3
Answers
GPT 3.5
To find the derivative of the function $f(x) = \left(3 + \cos^2(z)\right)^3$, we need to apply the chain rule.
Let $g(x) = 3 + \cos^2(z)$ and $h(x) = g(x)^3$. We are given $h(x) = f(x)$.
We can find the derivative of $g(x)$ first and then use it to find the derivative of $h(x)$.
To find the derivative of $g(x)$, we treat $z$ as a function of $x$ and apply the chain rule.
$\frac{dg}{dx} = \frac{d}{dx}\left(3 + \cos^2(z)\right) = 0 + \frac{d}{dz}\left(\cos^2(z)\right) \cdot \frac{dz}{dx}$
Using the chain rule, the derivative of $\cos^2(z)$ with respect to $z$ is $2\cos(z) \cdot \left(-\sin(z)\right) = -2\cos(z) \cdot \sin(z)$.
Therefore, $\frac{dg}{dx} = -2\cos(z) \cdot \sin(z) \cdot \frac{dz}{dx}$.
Now, to find the derivative of $h(x)$, we use the chain rule:
$\frac{dh}{dx} = \frac{d}{dx}\left(g(x)^3\right) = 3g(x)^2 \cdot \frac{dg}{dx}$.
Since we have already found $\frac{dg}{dx} = -2\cos(z) \cdot \sin(z) \cdot \frac{dz}{dx}$, we can substitute it into the formula for $\frac{dh}{dx}$:
$\frac{dh}{dx} = 3g(x)^2 \cdot \frac{dg}{dx} = 3\left(3 + \cos^2(z)\right)^2 \cdot (-2\cos(z) \cdot \sin(z) \cdot \frac{dz}{dx})$.
Therefore, the derivative of $f(x) = \left(3 + \cos^2(z)\right)^3$ with respect to $x$ is:
$\frac{d}{dx}\left(3 + \cos^2(z)\right)^3 = 3\left(3 + \cos^2(z)\right)^2 \cdot (-2\cos(z) \cdot \sin(z) \cdot \frac{dz}{dx})$.
Let $g(x) = 3 + \cos^2(z)$ and $h(x) = g(x)^3$. We are given $h(x) = f(x)$.
We can find the derivative of $g(x)$ first and then use it to find the derivative of $h(x)$.
To find the derivative of $g(x)$, we treat $z$ as a function of $x$ and apply the chain rule.
$\frac{dg}{dx} = \frac{d}{dx}\left(3 + \cos^2(z)\right) = 0 + \frac{d}{dz}\left(\cos^2(z)\right) \cdot \frac{dz}{dx}$
Using the chain rule, the derivative of $\cos^2(z)$ with respect to $z$ is $2\cos(z) \cdot \left(-\sin(z)\right) = -2\cos(z) \cdot \sin(z)$.
Therefore, $\frac{dg}{dx} = -2\cos(z) \cdot \sin(z) \cdot \frac{dz}{dx}$.
Now, to find the derivative of $h(x)$, we use the chain rule:
$\frac{dh}{dx} = \frac{d}{dx}\left(g(x)^3\right) = 3g(x)^2 \cdot \frac{dg}{dx}$.
Since we have already found $\frac{dg}{dx} = -2\cos(z) \cdot \sin(z) \cdot \frac{dz}{dx}$, we can substitute it into the formula for $\frac{dh}{dx}$:
$\frac{dh}{dx} = 3g(x)^2 \cdot \frac{dg}{dx} = 3\left(3 + \cos^2(z)\right)^2 \cdot (-2\cos(z) \cdot \sin(z) \cdot \frac{dz}{dx})$.
Therefore, the derivative of $f(x) = \left(3 + \cos^2(z)\right)^3$ with respect to $x$ is:
$\frac{d}{dx}\left(3 + \cos^2(z)\right)^3 = 3\left(3 + \cos^2(z)\right)^2 \cdot (-2\cos(z) \cdot \sin(z) \cdot \frac{dz}{dx})$.