let ( f ) be a differentiable function with ( f(-2)=-2,f(-2)=2,f(5)=-3,f(5)=5 ). let the function (…

Explanation:

Step1: Recall the chain - rule

The chain - rule states that if (y = f(u)) and (u = g(x)), then (\frac{dy}{dx}=f^{\prime}(u)g^{\prime}(x)). Here, (g(x)=2x^{2}-3), and we want to find the derivative of (y = f(g(x))) at (x = 2). First, find (g(2)) and (g^{\prime}(x)).

Step2: Calculate (g(2))

[ \begin{align*} g(x)&=2x^{2}-3\ g(2)&=2\times2^{2}-3\ &=2\times4 - 3\ &=8 - 3\ &=5 \end{align*} ]

Step3: Calculate (g^{\prime}(x))

Using the power - rule ((x^{n})^\prime=nx^{n - 1}), we have (g^{\prime}(x)=(2x^{2}-3)^\prime = 4x). Then (g^{\prime}(2)=4\times2 = 8).

Step4: Use the chain - rule

The derivative of (y = f(g(x))) with respect to (x) is ((f(g(x)))^\prime=f^{\prime}(g(x))g^{\prime}(x)). We want to find the value at (x = 2). Substitute (x = 2) into ((f(g(x)))^\prime): ((f(g(2)))^\prime=f^{\prime}(g(2))g^{\prime}(2)). Since (g(2)=5) and (g^{\prime}(2)=8), and (f^{\prime}(5)=-3), we have ((f(g(2)))^\prime=f^{\prime}(5)g^{\prime}(2)=(-3)\times8=-24).

The equation of the tangent line to (y = f(g(x))) at (x = 2) is of the form (y - y_0=m(x - x_0)), where (x_0 = 2), (y_0=f(g(2))=f(5)=2), and (m=(f(g(x)))^\prime|_{x = 2}=-24).

The equation of the tangent line is (y - 2=-24(x - 2)). Expand it: (y-2=-24x + 48), so (y=-24x+50).

Answer:

(y=-24x + 50)