Question

1. if $f(x) = \cosh 2x - \sinh 2x$, then $f^{(33)}(0) = $

(a) $2^{35}$
(b) $2^{33}$
(c) $-2^{35}$
(d) $-2^{33}$

Explanation:

Step1: Recall derivatives of hyperbolic functions

The derivative of \(\cosh(u)\) is \(\sinh(u) \cdot u'\) and the derivative of \(\sinh(u)\) is \(\cosh(u) \cdot u'\) by the chain rule. For \(f(x)=\cosh(2x)-\sinh(2x)\), let's find the first few derivatives to identify the pattern.

First derivative:
\(f'(x)=\frac{d}{dx}(\cosh(2x))-\frac{d}{dx}(\sinh(2x)) = 2\sinh(2x)-2\cosh(2x)\)

Second derivative:
\(f''(x)=2\times2\cosh(2x)-2\times2\sinh(2x)=4\cosh(2x)-4\sinh(2x)\)

Third derivative:
\(f'''(x)=4\times2\sinh(2x)-4\times2\cosh(2x)=8\sinh(2x)-8\cosh(2x)\)

Fourth derivative:
\(f^{(4)}(x)=8\times2\cosh(2x)-8\times2\sinh(2x)=16\cosh(2x)-16\sinh(2x)\)

We can observe a pattern here. For the \(n\)-th derivative:

• If \(n\) is odd: \(f^{(n)}(x)=2^{n}(\sinh(2x)-\cosh(2x))\)
• If \(n\) is even: \(f^{(n)}(x)=2^{n}(\cosh(2x)-\sinh(2x))\)

Step2: Determine the 33rd derivative

Since \(33\) is odd, we use the formula for odd \(n\): \(f^{(33)}(x)=2^{33}(\sinh(2x)-\cosh(2x))\)

Step3: Evaluate at \(x = 0\)

We know that \(\sinh(0)=0\) and \(\cosh(0)=1\). Substitute \(x = 0\) into \(f^{(33)}(x)\):

\(f^{(33)}(0)=2^{33}(\sinh(0)-\cosh(0))=2^{33}(0 - 1)=-2^{33}\) Wait, but let's re - check the pattern. Wait, when we took the first derivative (\(n = 1\), odd): \(f'(x)=2(\sinh(2x)-\cosh(2x))\), at \(x = 0\), \(f'(0)=2(0 - 1)=-2\)

Second derivative (\(n = 2\), even): \(f''(x)=4(\cosh(2x)-\sinh(2x))\), at \(x = 0\), \(f''(0)=4(1 - 0)=4 = 2^{2}\)

Third derivative (\(n=3\), odd): \(f'''(x)=8(\sinh(2x)-\cosh(2x))\), at \(x = 0\), \(f'''(0)=8(0 - 1)=-8=-2^{3}\)

Ah, so the general formula for \(f^{(n)}(0)\) is:

• If \(n\) is odd: \(f^{(n)}(0)=-2^{n}\)
• If \(n\) is even: \(f^{(n)}(0)=2^{n}\)

Since \(n = 33\) (odd), \(f^{(33)}(0)=-2^{33}\)? Wait, but the options have \(- 2^{33}\) as option (d). Wait, but let's re - derive the derivatives carefully.

Wait, \(f(x)=\cosh(2x)-\sinh(2x)\)

We know that \(\cosh(2x)-\sinh(2x)=e^{-2x}\) (from the definitions \(\cosh t=\frac{e^{t}+e^{-t}}{2}\), \(\sinh t=\frac{e^{t}-e^{-t}}{2}\), so \(\cosh(2x)-\sinh(2x)=\frac{e^{2x}+e^{-2x}}{2}-\frac{e^{2x}-e^{-2x}}{2}=\frac{e^{2x}+e^{-2x}-e^{2x}+e^{-2x}}{2}=e^{-2x}\))

Now, let's find the derivatives of \(y = e^{-2x}\)

First derivative: \(y'=-2e^{-2x}\)

Second derivative: \(y''=(-2)\times(-2)e^{-2x}=2^{2}e^{-2x}\)

Third derivative: \(y'''=2^{2}\times(-2)e^{-2x}=-2^{3}e^{-2x}\)

\(\cdots\)

The \(n\)-th derivative of \(y = e^{-2x}\) is \(y^{(n)}=(-2)^{n}e^{-2x}\)

Now, evaluate at \(x = 0\): \(y^{(n)}(0)=(-2)^{n}e^{0}=(-2)^{n}\)

Since \(f(x)=e^{-2x}\), then \(f^{(n)}(x)=(-2)^{n}e^{-2x}\), so \(f^{(n)}(0)=(-2)^{n}\)

For \(n = 33\), \(f^{(33)}(0)=(-2)^{33}=-2^{33}\)

Answer:

\(\boxed{-2^{33}}\) (which corresponds to option (d))