If $q(x)=\arcsin(x)+(2x-1)^3$, find the instantaneous rate of change of $q(x)$ at $x=0$

Key takeaway: This is a direct differentiation problem. The instantaneous rate of change at a point is the derivative evaluated at that x-value.

We want the instantaneous rate of change of

$q(x)=\arcsin(x)+(2x-1)^3$

at $x=0$, so we compute $q'(0)$.

Step 1: Differentiate each term

$\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}$

For the second term, use the chain rule:

$\frac{d}{dx}[(2x-1)^3] = 3(2x-1)^2 \cdot 2 = 6(2x-1)^2$

So,

$q'(x)=\frac{1}{\sqrt{1-x^2}}+6(2x-1)^2$

Step 2: Evaluate at $x=0$

$q'(0)=\frac{1}{\sqrt{1-0^2}}+6(2\cdot 0-1)^2$

$q'(0)=1+6(1)=7$

Answer

$\boxed{7}$

So the instantaneous rate of change of $q(x)$ at $x=0$ is $7$.

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Pitfalls the pros know 👇 A common mistake is forgetting the chain rule on $(2x-1)^3$. The inner derivative contributes a factor of 2, so the derivative is $6(2x-1)^2$, not $3(2x-1)^2$.

What if the problem changes? If the question asked for the instantaneous rate of change at another point, the method is the same: find $q'(x)$ first, then substitute the new x-value. If the domain were restricted, you would also need to check that the point lies inside the domain of $\arcsin(x)$.

Tags: derivative, chain rule, instantaneous rate of change