Let h(x) = arcsin(3x - 1). Find h'left(1/3right).

Chain rule with arcsine

Let $h(x) = \arcsin(3x - 1)$ . Find $h'\left(\frac{1}{3}\right)$ .

Hint

Use the chain rule with outer function $\arcsin(u)$ , whose derivative is $\frac{1}{\sqrt{1-u^2}}$ , and inner function $u = 3x - 1$ . Differentiate the inner function separately, then multiply the two derivatives together before plugging in $x = \frac{1}{3}$ .

Step-by-step solution

Identify the outer function $f(u) = \arcsin(u)$ with $f'(u) = \frac{1}{\sqrt{1-u^2}}$ , and the inner function $g(x) = 3x - 1$ with $g'(x) = 3$ . By the chain rule, $h'(x) = f'(g(x)) \cdot g'(x) = \frac{3}{\sqrt{1 - (3x-1)^2}}$ .

Now substitute $x = \frac{1}{3}$ : since $3(1/3) - 1 = 0$ , we get $h'(1/3) = \frac{3}{\sqrt{1 - 0^2}} = \frac{3}{1} = 3$ .

Chain rule with arcsine

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