The derivative of the function y = cot^(-1)(vZx) is: (20+2)v(2x+3) (26-1)v(2x-3) This option This option y’ y’ (20+2)v(2x+3) (26-1)v(2x-3) This option This option
The derivative of the function y = cot^(-1)(vZx) is: (20+2)v(2x+3) (26-1)v(2x-3) This option This option y’ y’ (20+2)v(2x+3) (26-1)v(2x-3) This option This option
The Correct Answer and Explanation is:
To find the derivative of y=cot−1(vZx)y = \cot^{-1}(vZx)y=cot−1(vZx), where vZxvZxvZx is the expression inside the inverse cotangent function, we need to apply the chain rule and the derivative formula for the inverse cotangent function.
Step 1: Apply the derivative rule for cot−1(x)\cot^{-1}(x)cot−1(x)
The derivative of the inverse cotangent function, cot−1(x)\cot^{-1}(x)cot−1(x), is given by:ddxcot−1(x)=−11+x2\frac{d}{dx} \cot^{-1}(x) = -\frac{1}{1 + x^2}dxdcot−1(x)=−1+x21
Now, in your case, the function inside the inverse cotangent is vZxvZxvZx. So, the derivative of y=cot−1(vZx)y = \cot^{-1}(vZx)y=cot−1(vZx) with respect to xxx is:y′=−ddx[cot−1(vZx)]y’ = -\frac{d}{dx} \left[ \cot^{-1}(vZx) \right]y′=−dxd[cot−1(vZx)]
Step 2: Apply the chain rule
We need to apply the chain rule because vZxvZxvZx is a function of xxx. To differentiate it, we first find the derivative of the inside function vZxvZxvZx with respect to xxx. If vZxvZxvZx is of the form v⋅Zxv \cdot Zxv⋅Zx, then the derivative of vZxvZxvZx with respect to xxx is simply vZvZvZ.
Thus, applying the chain rule:y′=−11+(vZx)2⋅ddx(vZx)y’ = -\frac{1}{1 + (vZx)^2} \cdot \frac{d}{dx}(vZx)y′=−1+(vZx)21⋅dxd(vZx)
The derivative of vZxvZxvZx with respect to xxx is vZvZvZ, so:y′=−vZ1+(vZx)2y’ = -\frac{vZ}{1 + (vZx)^2}y′=−1+(vZx)2vZ
Step 3: Simplify and Interpret the Options
You mentioned several options like:
- (20+2)v(2x+3)
- (26-1)v(2x-3)
These options don’t match the standard form of the derivative for cot−1(vZx)\cot^{-1}(vZx)cot−1(vZx), as they involve linear terms in xxx and constants that don’t appear in the general derivative formula for inverse cotangent. It’s likely that there is a specific context or additional information about the function vZxvZxvZx that has been omitted. However, based on the standard derivative rules, the correct expression for the derivative would be:y′=−vZ1+(vZx)2y’ = -\frac{vZ}{1 + (vZx)^2}y′=−1+(vZx)2vZ
