Lse an appropriate method of
ifferentiation determine
ne following

Ask by Harris Li. in South Africa

Jan 22,2025

Question

Lse an appropriate method of ifferentiation determine ne following $$ f(x)=\cos (\sqrt{\sin (\tan \pi x)}) $$

UpStudy AI · Accepted Answer

To differentiate the function
$$
f(x) = \cos\left( \sqrt{ \sin\left( \tan(\pi x) \right) } \right),
$$
we can apply the **chain rule** multiple times due to the nested functions involved. Here's a step-by-step breakdown:

### Step 1: Identify the Nested Functions
Let's break down $$ f(x) $$ into simpler functions:

1. **Innermost function:**
   $$
   u = \tan(\pi x)
   $$
2. **Next layer:**
   $$
   v = \sin(u) = \sin(\tan(\pi x))
   $$
3. **Following layer:**
   $$
   w = \sqrt{v} = \sqrt{\sin(\tan(\pi x))}
   $$
4. **Outer function:**
   $$
   f(x) = \cos(w) = \cos\left( \sqrt{\sin(\tan(\pi x))} \right)
   $$

### Step 2: Apply the Chain Rule
To find $$ f'(x) $$, we'll differentiate each layer step by step.

1. **Differentiate $$ f $$ with respect to $$ w $$:**
   $$
   \frac{df}{dw} = -\sin(w)
   $$
   
2. **Differentiate $$ w $$ with respect to $$ v $$:**
   $$
   \frac{dw}{dv} = \frac{1}{2\sqrt{v}} = \frac{1}{2\sqrt{\sin(\tan(\pi x))}}
   $$
   
3. **Differentiate $$ v $$ with respect to $$ u $$:**
   $$
   \frac{dv}{du} = \cos(u) = \cos(\tan(\pi x))
   $$
   
4. **Differentiate $$ u $$ with respect to $$ x $$:**
   $$
   \frac{du}{dx} = \frac{d}{dx} [\tan(\pi x)] = \pi \sec^2(\pi x)
   $$

### Step 3: Combine the Derivatives
Using the chain rule, multiply the derivatives from each layer:

$$
f'(x) = \frac{df}{dw} \cdot \frac{dw}{dv} \cdot \frac{dv}{du} \cdot \frac{du}{dx}
$$

Substituting the derivatives:

$$
f'(x) = \left[ -\sin\left( \sqrt{\sin(\tan(\pi x))} \right) \right] \cdot \left[ \frac{1}{2\sqrt{\sin(\tan(\pi x))}} \right] \cdot \cos(\tan(\pi x)) \cdot \pi \sec^2(\pi x)
$$

### Step 4: Simplify the Expression
Combining the constants and trigonometric functions, the derivative simplifies to:

$$
f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^2(\pi x)}{2 \sqrt{\sin(\tan(\pi x))}}
$$

Alternatively, it can be written as:

$$
f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x))}{2 \cos^2(\pi x) \sqrt{\sin(\tan(\pi x))}}
$$

### Final Answer
Thus, the derivative of $$ f(x) $$ is:

$$
f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^2(\pi x)}{2 \sqrt{\sin(\tan(\pi x))}}
$$

### Summary
By applying the chain rule methodically to each nested function, we successfully derived the derivative of the given complex trigonometric function.
To find the derivative of $$ f(x) = \cos\left( \sqrt{ \sin(\tan(\pi x)) } \right) $$, use the chain rule multiple times: 1. Differentiate the outer cosine function: $$ -\sin(w) $$. 2. Differentiate the square root: $$ \frac{1}{2\sqrt{v}} $$. 3. Differentiate the sine function: $$ \cos(u) $$. 4. Differentiate the tangent function: $$ \pi \sec^2(\pi x) $$. Combine these derivatives: $$ f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^2(\pi x)}{2 \sqrt{\sin(\tan(\pi x))}} $$ So, the derivative is: $$ f'(x) = -\frac{\pi \sin\left( \sqrt{\sin(\tan(\pi x))} \right) \cos(\tan(\pi x)) \sec^2(\pi x)}{2 \sqrt{\sin(\tan(\pi x))}} $$

Lse an appropriate method of
ifferentiation determine
ne following

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