CBSE Class 12 Mathematics Derivatives Q #1444
COMPETENCY BASED
APPLY
2 Marks 2025 AISSCE(Board Exam) VSA
Differentiate $\sqrt{e^{\sqrt{2x}}}$ with respect to $e^{\sqrt{2x}}$ for $x>0$.

AI Tutor Explanation

Powered by Gemini

Detailed Solution

Step 1: Define the functions

Let $u = \sqrt{e^{\sqrt{2x}}}$ and $v = e^{\sqrt{2x}}$. We need to find $\frac{du}{dv}$.

Step 2: Express u in terms of v

We have $u = \sqrt{v}$, since $u = \sqrt{e^{\sqrt{2x}}} = \sqrt{v}$.

Step 3: Differentiate u with respect to v

Now, we differentiate $u$ with respect to $v$: $$ \frac{du}{dv} = \frac{d}{dv}(\sqrt{v}) = \frac{d}{dv}(v^{\frac{1}{2}}) = \frac{1}{2}v^{\frac{1}{2} - 1} = \frac{1}{2}v^{-\frac{1}{2}} = \frac{1}{2\sqrt{v}} $$

Step 4: Substitute the value of v

Substitute $v = e^{\sqrt{2x}}$ back into the expression: $$ \frac{du}{dv} = \frac{1}{2\sqrt{e^{\sqrt{2x}}}} $$

Step 5: Simplify the expression

We can simplify the expression further: $$ \frac{du}{dv} = \frac{1}{2\sqrt{e^{\sqrt{2x}}}} = \frac{1}{2e^{\frac{\sqrt{2x}}{2}}} $$

Final Answer: $\frac{1}{2e^{\frac{\sqrt{2x}}{2}}}$

AI generated content. Review strictly for academic accuracy.

Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student needs to apply the chain rule and differentiation formulas to solve the problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires the student to apply the rules of differentiation, specifically the chain rule, to find the derivative. This involves a series of steps and procedures.
Syllabus Audit: In the context of CBSE Class 12, this is classified as COMPETENCY. The question assesses the student's ability to apply differentiation techniques in a non-standard way, requiring them to differentiate with respect to another function rather than a variable.

More from this Chapter

SUBJECTIVE
Amusement parks heavily rely on mathematical curves to design roller coaster tracks, balancing intense thrill with absolute safety. For a safe ride, the track must have no breaks (it must be continuous) and no sharp, sudden corners (it must be differentiable/smooth). An engineering team is designing a transition segment of a new coaster. The height $h(x)$ (in meters) of the track at a horizontal distance $x$ (in meters) from the starting platform is modeled by the following piecewise function: $$h(x) = \begin{cases} \frac{1}{2}x^2 + 2x, & 0 \leq x < 2 \\ ax + b, & 2 \leq x \leq 5 \end{cases}$$ Based on the given information, answer the following questions: (i) For the track to have no broken rails at the transition point $x = 2$, it must be continuous. Formulate the mathematical equation relating $a$ and $b$ to ensure this continuity. (1 Mark) (ii) Find the left-hand derivative (the slope of the track just before the transition point) at $x = 2$. (1 Mark) (iii) To ensure the transition at $x = 2$ is perfectly smooth (differentiable), calculate the exact values of $a$ and $b$ that the engineers must use. (2 Marks) — OR — (iii) If a junior engineer mistakenly designed the second track segment using $a = 3$ and $b = 0$, verify whether the track is mathematically safe (smooth and differentiable) at $x = 2$. Justify your answer. (2 Marks)
LA
Find the differential of $x^{\cot x}+\frac{2x^{2}-3}{2x^{2}-x+2}$ with respect to x.
VSA
If $x=a\sin^{3}t$, $y=b\cos^{3}t$, then find $\frac{dy}{dx}$ at $t=\frac{\pi}{4}$.
LA
If $y\sqrt{x^{2}+1}=\log\sqrt{x^{2}+1}-x$, show that $(x^{2}+1)\frac{dy}{dx}+xy+1=0$.
LA
Find $\frac{dy}{dx}$ if $y^{x}+x^{y}+x^{x}=a^{b}$, where a and b are constants.
View All Questions