Question

Let f(x) = x3 + x + 1. Which expression gives StartFraction d Over d x EndFraction (f Superscript negative 1 Baseline (x) )?

StartFraction 1 Over f Superscript negative 1 Baseline (x) EndFraction

StartFraction 1 Over f Superscript negative 1 Baseline (3 x squared + 1) EndFraction

StartFraction 1 Over (3 x squared + 1) f Superscript negative 1 Baseline (x) EndFraction

StartFraction 1 Over 3 (F Superscript negative 1 Baseline (x) ) squared + 1 EndFraction

Answer 1

To find the expression for StartFraction d Over d x EndFraction (f Superscript negative 1 Baseline (x)), we can use the chain rule of differentiation. The expression is 1 / (3x^2 + 1).

Step-by-step explanation:

To find the expression for StartFraction d Over d x EndFraction (f Superscript negative 1 Baseline (x)), we can use the chain rule of differentiation. Let's denote f inverse (x) as g(x). We want to find StartFraction d Over d x EndFraction (g(x)).

Using the chain rule, we have:

StartFraction d Over d x EndFraction (g(x)) = StartFraction d Over d x EndFraction (f inverse (x)) = 1 / (StartFraction d Over d f EndFraction (f inverse (x))).

Now, let's differentiate f(x) = x3 + x + 1.

StartFraction d Over d f EndFraction (x) = 3x^2 + 1.

Substituting this into the expression, we get:

StartFraction d Over d x EndFraction (f Superscript negative 1 Baseline (x)) = 1 / (3x^2 + 1).