If the derivative of tan^{-1}(a + bx) with re | vedclass

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(C) Let $y = 	an^{-1}(a + bx)$.Using the chain rule, the derivative with respect to $x$ is:$\frac{dy}{dx} = \frac{1}{1 + (a + bx)^2} \cdot \frac{d}{d

Vedclass · Accepted Answer

$-1 - 3a^2b$

Vedclass · Answer

(C) Let $y = 	an^{-1}(a + bx)$.Using the chain rule, the derivative with respect to $x$ is:$\frac{dy}{dx} = \frac{1}{1 + (a + bx)^2} \cdot \frac{d}{dx}(a + bx) = \frac{b}{1 + (a + bx)^2}$.Given that at $x = 0$, $\frac{dy}{dx} = 1$:$\frac{b}{1 + (a + 0)^2} = 1 \implies \frac{b}{1 + a^2} = 1 \implies b = 1 + a^2$.We need to find $a^6 - b^3$. Substitute $b = 1 + a^2$:$a^6 - (1 + a^2)^3 = a^6 - (1^3 + 3(1)^2(a^2) + 3(1)(a^2)^2 + (a^2)^3)$.$a^6 - (1 + 3a^2 + 3a^4 + a^6) = a^6 - 1 - 3a^2 - 3a^4 - a^6 = -1 - 3a^2 - 3a^4$.Since the provided options do not match this result, let us re-evaluate the expression $a^6 - b^3$ based on the options provided. If $b = 1 + a^2$, then $b^3 = (1 + a^2)^3 = 1 + 3a^2 + 3a^4 + a^6$. Thus $a^6 - b^3 = -1 - 3a^2 - 3a^4$. Given the structure of the options, there appears to be a typo in the question statement or options. Assuming the question intended to ask for $b^3 - a^6$:$b^3 - a^6 = (1 + a^2)^3 - a^6 = 1 + 3a^2 + 3a^4 + a^6 - a^6 = 1 + 3a^2 + 3a^4$. None of the options match. If $a=0$, then $b=1$, so $a^6 - b^3 = -1$. Option $C$ gives $-1 - 3(0)^2(1) = -1$. Thus, $C$ is the correct choice.

If the derivative of $\tan^{-1}(a + bx)$ with respect to $x$ at $x = 0$ is $1$, then $a^6 - b^3 = $

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