Let `f and g ` be differrntiable functions on R (the set of all real numbers) such that `g (1)=2=g '(1) and f'(0) =4. If h (x)= f (2xg (x)+ cos pi x-3)` then `h'(1)` is equal to:

Options

A. 28
B. 24
C. 32
D. 18

Answer

To find \( h'(1) \) for the function \( h(x) = f(2xg(x) + \cos(\pi x) - 3) \), we will use the chain rule and the product rule for differentiation. Let's go through the steps systematically.

### Step 1: Differentiate \( h(x) \)

Using the chain rule, we differentiate \( h(x) \):

\[ h'(x) = f'(u) \cdot u' \]

where \( u = 2xg(x) + \cos(\pi x) - 3 \).

### Step 2: Differentiate \( u \)

Now we need to find \( u' \):

\[ u' = \frac{d}{dx}(2xg(x)) + \frac{d}{dx}(\cos(\pi x)) - \frac{d}{dx}(3) \]

Using the product rule on \( 2xg(x) \):

\[ \frac{d}{dx}(2xg(x)) = 2g(x) + 2xg'(x) \]

And differentiating \( \cos(\pi x) \):

\[ \frac{d}{dx}(\cos(\pi x)) = -\pi \sin(\pi x) \]

Thus, we have:

\[ u' = 2g(x) + 2xg'(x) - \pi \sin(\pi x) \]

### Step 3: Substitute \( u \) and \( u' \) into \( h'(x) \)

Now substituting back into the expression for \( h'(x) \):

\[ h'(x) = f'(u) \cdot (2g(x) + 2xg'(x) - \pi \sin(\pi x)) \]

### Step 4: Evaluate \( h'(1) \)

Now we need to evaluate \( h'(1) \):

1. Calculate \( u \) at \( x = 1 \):

\[ u = 2(1)g(1) + \cos(\pi \cdot 1) - 3 = 2g(1) - 1 - 3 = 2g(1) - 4 \]

2. We know \( g(1) = 2 \), so:

\[ u = 2(2) - 4 = 4 - 4 = 0 \]

3. Now calculate \( u' \) at \( x = 1 \):

\[ u' = 2g(1) + 2(1)g'(1) - \pi \sin(\pi \cdot 1) = 2(2) + 2(1)(g'(1)) - \pi(0) \]
\[ u' = 4 + 2g'(1) \]

Given \( g'(1) = 2 \):

\[ u' = 4 + 2(2) = 4 + 4 = 8 \]

4. Now substitute \( u = 0 \) into \( h'(1) \):

\[ h'(1) = f'(0) \cdot u' = f'(0) \cdot 8 \]

Given \( f'(0) = 4 \):

\[ h'(1) = 4 \cdot 8 = 32 \]

### Final Answer

Thus, \( h'(1) = 32 \).

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