Answer:
The answer is option a
Step-by-step explanation:
Given,
f(x) = [tex]\sqrt{x^{2}-4 }[/tex]
g(x) = 3x - 2
By substituting g(x) in f(x) we get
f(g(x)) = [tex]\sqrt{(3x-2)^{2} -4}[/tex]
By differentiating with respect to we get
[tex]\frac{\mathrm{d}\sqrt{(3x-2)^{2} -4}}{\mathrm{d} x}[/tex] = [tex]1/2\sqrt{(3x-2)^{2}-4} [/tex]×2×(3x-2)×3
By using chain rule we get this equation
[tex]{\mathrm{d}f(g(x))}/{\mathrm{d} x}[/tex] = [tex]1/2\sqrt{(3x-2)^{2} -4} [/tex]×2×(3x-2)×3
Substituting x = 3 we get
[tex]{\mathrm{d}f(g(x))}/{\mathrm{d} x}[/tex] = 7/[tex]\sqrt{5}[/tex]
Answer is option a
Answer:
OPTION A: [tex]$ \frac{7}{\sqrt{5}} $[/tex]
Step-by-step explanation:
Given: f(x) = [tex]$ \sqrt{x^2 - 4} $[/tex] and g(x) = 3x - 2
We are to find f(g(x)).
f(g(x)) = f(3x - 2)
Substituting 3x - 2 in place of x in f(x), we get
f(g(x)) = [tex]$ \sqrt{(3x - 2)^2 - 4}$[/tex]
[tex]$ \implies \sqrt{9x^ 2 - 12x + 4 - 4} = \sqrt{9x^2 -12x} $[/tex]
Differentiating this we get:
[tex]$ f(g(x))^' = \frac{1}{2} (9x^2 - 12x)^{\frac{-1}{2}}} \times (18x - 12) $[/tex]
At x = 3,
[tex]$ f(g(3))^' = \frac{1}{2}[9(3)^2 - 12(3)]^{\frac{-1}{2}}} \times [18(3) - 12] $[/tex]
[tex]$ = \frac{1}{2} [81 - 36]^{\frac{1}{2}}} \times 42 $[/tex]
[tex]$ = \frac{21}{\sqrt{45}} $[/tex]
[tex]$ \frac{21}{3\sqrt{5}} $[/tex]
[tex]$ \frac{7}{\sqrt{5}}} $[/tex]
Therefore, the answer is 7/√5.
