If `y=f(x^(3)),z=g(x^(2)),f'(x),=cosx` and `g'(x)=sinx," then "(dy)/(dz)` is

Options

A. `(3x)/(2)cosx^(3)cosecx^(2)`
B. `(2)/(3)sinx^(3)secx^(2)`
C. `tanx`
D. none of these

Explanation

To find \(\frac{dy}{dz}\) given \(y = f(x^3)\) and \(z = g(x^2)\), where \(f'(x) = \cos x\) and \(g'(x) = \sin x\), we will follow these steps:

### Step 1: Differentiate \(y\) with respect to \(x\)
We start with the function:
\[ y = f(x^3) \]
Using the chain rule, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = f'(x^3) \cdot \frac{d}{dx}(x^3) \]
The derivative of \(x^3\) is \(3x^2\), so we have:
\[ \frac{dy}{dx} = f'(x^3) \cdot 3x^2 \]

### Step 2: Substitute \(f'(x^3)\)
We know that \(f'(x) = \cos x\). Therefore, substituting \(x^3\) into this gives:
\[ f'(x^3) = \cos(x^3) \]
Now we can substitute this back into our equation for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = 3x^2 \cdot \cos(x^3) \]

### Step 3: Differentiate \(z\) with respect to \(x\)
Now we differentiate \(z\):
\[ z = g(x^2) \]
Using the chain rule again, we differentiate \(z\) with respect to \(x\):
\[ \frac{dz}{dx} = g'(x^2) \cdot \frac{d}{dx}(x^2) \]
The derivative of \(x^2\) is \(2x\), so we have:
\[ \frac{dz}{dx} = g'(x^2) \cdot 2x \]

### Step 4: Substitute \(g'(x^2)\)
We know that \(g'(x) = \sin x\). Therefore, substituting \(x^2\) into this gives:
\[ g'(x^2) = \sin(x^2) \]
Now we can substitute this back into our equation for \(\frac{dz}{dx}\):
\[ \frac{dz}{dx} = 2x \cdot \sin(x^2) \]

### Step 5: Find \(\frac{dy}{dz}\)
Now we can find \(\frac{dy}{dz}\) using the chain rule:
\[ \frac{dy}{dz} = \frac{dy/dx}{dz/dx} \]
Substituting the expressions we found:
\[ \frac{dy}{dz} = \frac{3x^2 \cdot \cos(x^3)}{2x \cdot \sin(x^2)} \]
We can simplify this:
\[ \frac{dy}{dz} = \frac{3}{2} \cdot \frac{x^2}{x} \cdot \frac{\cos(x^3)}{\sin(x^2)} = \frac{3}{2} \cdot x \cdot \frac{\cos(x^3)}{\sin(x^2)} \]

### Final Answer
Thus, the final expression for \(\frac{dy}{dz}\) is:
\[ \frac{dy}{dz} = \frac{3}{2} x \cdot \cot(x^2) \]