The function \(y=sin⁡(x^{2})\)contains an outer trigonometric function and an inner polynomial function. Using the chain rule: \(\frac{d}{dx}[sin⁡(x^{2})]=cos⁡(x^{2})⋅2x\) Option C is correct because both the derivative of \(\sin\,u\) and the derivative of \(u=x^{2}\)are included. Option A ignores the inner derivative, Option B uses the wrong trigonometric function, and Option D incorrectly simplifies the expression.

The extended chain rule applies when variables depend successively on one another. If \(y\) depends on \(u\), \(u\) on \(v\), and \(v\) on \(x\), then derivatives multiply successively: \(\frac{dy}{dx}=\frac{dy}{du}⋅\frac{du}{dv}⋅\frac{dv}{dx}\) The rule works only when all intermediate derivatives exist. Therefore, both Assertion and Reason are true, and the Reason correctly explains the Assertion.

Both given functions are compositions of inner polynomial expressions and outer power functions. Using the chain rule: \(\frac{d}{dx}(x^{2}+1)^{5}=5(x^{2}+1)^{4}⋅2x\) and \(\frac{d}{dx}(x^{3}+2x)^{4}=4(x^{3}+2x)^{3}(3x^{2}+2)\) Statement I is true because polynomial composition requires chain rule application. Hence all statements are correct.

Using chain rule: \(\frac{d}{dx}[sin⁡(x^{2})]=2xcos⁡(x^{2})\frac{d}{dx}[cos⁡(3x)]=-3sin⁡(3x)\frac{d}{dx}[tan⁡(x^{3})]=3x^{2}{sec⁡}^{2}(x^{3})\) Thus matching becomes \(1-a,2-b,3-c\). Option A correctly pairs each function with its derivative.

The correct differentiation order is: 1. Convert common logarithm using \(log⁡a=\frac{\ln\,a}{\ln\,10}\). 2. Take constant \(\frac{1}{\ln\,10}\)outside differentiation. 3. Differentiate \(ln⁡(2x^{2}+1)\)using chain rule. 4. Simplify the final expression. Therefore, the correct sequence is \(1,2,3,4\). The provided answer B is incorrect.

Vector differentiation is performed component-wise: \(\frac{d}{dt}\left[(t^{2})\hat{i}+(sin⁡t)\hat{j}\right]=2t\hat{i}+cos⁡t\hat{j}\) Option C correctly differentiates both components. Options A and B leave one component undifferentiated, while Option D interchanges derivatives incorrectly.

Implicit differentiation is used when variables are mixed in one equation and isolating \(y\) is difficult or inconvenient. In \(x^{2}+y^{2}=1\), differentiating directly gives the slope relation efficiently. Option D correctly explains the need for implicit differentiation. Other options do not describe the actual mathematical reason.

Differentiating implicitly: \(\frac{d}{dx}(x^{2}+xy+y^{2})=0\) gives \(2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0\) Collecting derivative terms: \(\left(x+2y)\frac{dy}{dx}=-(2x+y\right)\) Thus, \(\frac{dy}{dx}=\frac{-2x-y}{x+2y}\) Hence Option A is correct.

Differentiating implicitly: \(2x+2y\frac{dy}{dx}=0\) Solving: \(2y\frac{dy}{dx}=-2x\frac{dy}{dx}=-\frac{x}{y}\) Thus Option B is correct. The negative sign appears due to transposition after differentiation.

From \(xy=6\): \(x\frac{dy}{dx}+y=0\) So, \(\frac{dy}{dx}=-\frac{y}{x}\) When \(x=2\), using \(xy=6\), we get \(y=3\). Hence: \(\frac{dy}{dx}=-\frac{3}{2}\) Thus the correct value is \(-3/2\). Option labels repeat the same value, but Option A is selected first correctly.

Differentiating implicitly: \(\frac{d}{dx}(2x+3y)=0\) gives: \(2+3\frac{dy}{dx}=0\) So, \(\frac{dy}{dx}=-\frac{2}{3}\) Option A is correct. Since the equation is linear, the slope remains constant throughout the graph.

Differentiating implicitly: \(\frac{d}{dx}[sin⁡(xy)]=cos⁡(xy)⋅\frac{d}{dx}(xy)\) Using the product rule: \(\frac{d}{dx}(xy)=x\frac{dy}{dx}+y\) Hence, \(cos⁡(xy)(x\frac{dy}{dx}+y)=1\) Option B is incorrect because it ignores differentiation of the product \(xy\). Therefore, only \(dy/dx\) is insufficient.

Given \(sin⁡y=x\), using the identity: \({sin⁡}^{2}y+{cos⁡}^{2}y=1\) \(\theta\) \({sin⁡}^{2}\theta \approx 0.329,  {cos⁡}^{2}\theta \approx 0.671\) \({sin⁡}^{2}\theta +{cos⁡}^{2}\theta \approx 1\) θ = 35°|cos θ| = 0.819|sin θ| = 0.574cos² θsin² θ0.671 + 0.329 = 1 we obtain: \(x^{2}+{cos⁡}^{2}y=1\) Thus, \(cos⁡y=\sqrt{1-x^{2}}\) Substituting into the derivative formula gives: \(\frac{d}{dx}({sin⁡}^{-1}x)=\frac{1}{\sqrt{1-x^{2}}}\) Hence Option A is correct.

The passage clearly states that the domain of the exponential function \(y=a^{x}\)is all real numbers: \(\left(-\infty ,\infty \right)\) This means every real value of \(x\) is allowed. Option B directly matches the statement in the passage. The range, not domain, is restricted to positive real numbers.

The passage explicitly defines the natural exponential function as: \(y=e^{x}\) Hence its base is \(e\), known as Euler's number. Option C correctly identifies the special base associated with natural logarithms and exponential functions.

For \(y={tan⁡}^{-1}x\), let \(tan⁡y=x\). Differentiating: \({sec⁡}^{2}y\frac{dy}{dx}=1\) Using \({sec⁡}^{2}y=1+{tan⁡}^{2}y\) and \(tan⁡y=x\): \(\frac{d}{dx}({tan⁡}^{-1}x)=\frac{1}{1+x^{2}}\) Thus Option D is correct. The derivative is always positive.

The logarithmic function \(\ln\,x\) is defined only for positive real numbers. Therefore, logarithmic differentiation requires: \(x>0\) Option A correctly gives the domain condition. Negative numbers and zero are not included in the logarithm's domain.

The natural logarithm function exists only when its argument is positive: \(x>0\) Hence Option B correctly states the domain. Values equal to zero or less than zero are invalid inputs for logarithmic functions.

Given: \(y=x^{x}\) Taking logarithm: \(ln⁡y=xln⁡x\) Differentiating: \(\frac{1}{y}\frac{dy}{dx}=1+ln⁡x\) Multiplying by \(y=x^{x}\): \(\frac{d}{dx}(x^{x})=x^{x}(1+ln⁡x)\) Thus Option C is correct.

For logarithmic differentiation: \(\frac{d}{dx}[ln⁡(x^{2}+1)]=\frac{1}{x^{2}+1}⋅2x\) Thus, \(\frac{dy}{dx}=\frac{2x}{x^{2}+1}\) Option B is correct. Option D is numerically identical but duplicated. First correct occurrence is selected.