Question

Suppose \(f\) is differentiable for all real numbers with \(f(0)=-3, f(1)=3, f^{\prime}(0)=3,\) and \(f^{\prime}(1)=5 .\) Let \(g(x)=\sin (\pi f(x)) .\) Evaluate the following expressions. a. \(g^{\prime}(0)\) b. \(g^{\prime}(1)\)

Short answer

Question: Find the derivatives of the composite function \(g(x) = \sin(\pi f(x))\) at the points \(x = 0\) and \(x = 1\), where \(f(0) = -3\), \(f(1) = 3\), \(f'(0) = 3\), and \(f'(1) = 5\). Answer: a. \(g'(0) = -3\pi\) b. \(g'(1) = -5\pi\)

Step-by-step solution

Find the derivative of g(x)

We first need to find the derivative of \(g(x)\). Since g(x) is a composite function, we use the Chain Rule: \[g'(x) = \frac{dg}{dx} = \frac{dg}{df} \cdot \frac{df}{dx}\] We know that \(g(x) = \sin(\pi f(x))\), and we want to differentiate it with respect to \(f(x)\): \[\frac{dg}{df} = \frac{d}{df} \sin(\pi f(x)) = \pi \cos(\pi f(x))\] Now we have the expression for the derivative of g with respect to f.

Use the Chain Rule to find g'(x)

Now we apply the Chain Rule to find the derivative of g(x) with respect to x: \[g'(x) = \frac{dg}{dx} = \frac{dg}{df} \cdot \frac{df}{dx} = \pi \cos(\pi f(x)) \cdot f'(x)\]

Evaluate g'(0)

To find \(g'(0)\), we substitute \(x = 0\) into our expression for \(g'(x)\): \[g'(0) = \pi \cos(\pi f(0)) \cdot f'(0)\] We are given that \(f(0) = -3\) and \(f'(0) = 3\): \[g'(0) = \pi \cos(\pi (-3)) \cdot 3\] Since \(\cos(\pi x)\) is an even function, \(\cos(\pi (-3)) = \cos(3\pi) = -1\): \[g'(0) = \pi (-1) \cdot 3 = -3\pi\] So, \(g'(0) = -3\pi\).

Evaluate g'(1)

Similarly, to find \(g'(1)\), we substitute \(x = 1\) into our expression for \(g'(x)\): \[g'(1) = \pi \cos(\pi f(1)) \cdot f'(1)\] We are given that \(f(1) = 3\) and \(f'(1) = 5\): \[g'(1) = \pi \cos(\pi (3)) \cdot 5\] Again, since \(\cos(\pi x)\) is an even function, \(\cos(\pi (3)) = \cos(3\pi) = -1\): \[g'(1) = \pi (-1) \cdot 5 = -5\pi\] So, \(g'(1) = -5\pi\). In conclusion, we have: a. \(g^{\prime}(0) = -3\pi\) b. \(g^{\prime}(1) = -5\pi\)