Application of Derivatives
17 Board Maths previous year questions on Application of Derivatives — options free on every question; 2 include the answer & explanation free, the rest unlock with PYQ Pass.
Derivative of \(\cos \left(\sqrt{x}\right)\) is :
\(-\frac{\sin \left(\sqrt{x}\right)}{2\sqrt{x}}\)
\(\text{ We need to differentiate }\cos (\sqrt{x})\text{ with respect to }x\text{. }\)
Step 1: Apply Chain Rule
Let: \(y=\cos \left(\sqrt{x}\right)\)
\(\text{ Define }u=\sqrt{x}\Rightarrow {x}^{1/2}\text{, so that: }\\ y=\cos \left(u\right)\)
Now, differentiate both sides:
\(\frac{dy}{dx}=\frac{d}{du}\cos (u)⋅\frac{du}{dx}\)
Step 2: Compute the Derivatives
\(\frac{d(\cos u)}{du}=-\sin (u)\\ \frac{du}{dx}=\frac{d\left({x}^{(\frac{1}{2})}\right)}{dx}\)
Step 3: Multiply the Terms
\(\frac{dy}{dx}=-\frac{\sin \sqrt{x}}{2\sqrt{x}}\)
The function \(f(x)=x^3-3 x^2+12 x-18\) is :
strictly increasing on \(R\)
Given: \(f(x)={x}^{3}-3{x}^{2}+12x-18\)
Then, \({f}^{'}(x)=3{x}^{2}-6x+12=3\left({x}^{2}-2x+4\right)\)
since, \({x}^{2}-2x+4=(x-1{)}^{2}+3>0\forall x\in \mathrm{ℝ}\)
\(\Rightarrow {f}^{'}(x)>0\forall x\in \mathrm{ℝ}\)
\(\Rightarrow f\) is strictly increasing on \(\mathbb{R}\)
A cylindrical tank of radius 10 cm is being filled with sugar at the rate of \(100\pi {\mathrm{cm}}^{3}/\mathrm{s}\). The rate, at which the height of the sugar inside the tank is increasing, is :
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The function \(\mathrm{f}(x)=\frac{x}{2}+\frac{2}{x}\) has a local minima at \(x\) equal to :
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A cylindrical tank of radius \(10\mathrm{cm}\) is being filled with sugar at the rate of \(100\pi {\mathrm{cm}}^{3}/\mathrm{s}\). The rate, at which the height of the sugar inside the tank is increasing, is :
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The function \(\mathrm{f}(x)=\frac{x}{2}+\frac{2}{x}\) has a local minima at \(x\) equal to :
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Given a curve \(\mathrm{y}=7x-{x}^{3}\) and \(x\) increases at the rate of \(2\) units per second. The rate at which the slope of the curve is changing, when \(\mathrm{x}=5\) is:
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Let \(f\left(x\right)={x}^{2025}−{x}^{2000},x\in \left[0,1\right]\) and the minimum value of the function \(f\left(x\right)\) in the interval \(\left[0,1\right]\) be \({(80)}^{80}{(n)}^{−81}\). Then \(n\) is equal to
[JEE Main 2026, 22 Jan (Shift 1)]
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The function \(\mathrm{f}(x)={x}^{2}-4x+6\) is increasing in the interval:
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The function \(f(x)=x^3-3 x^2+12 x-18\) is :
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Let \(f(x)\) be a continuous function on \([a, b]\) and differentiable on \((a, b)\). Then, this function \(f(x)\) is strictly increasing in \((a, b)\) if
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Given a curve \(\mathrm{y}=7x-{x}^{3}\) and \(x\) increases at the rate of \(2\) units per second. The rate at which the slope of the curve is changing, when \(\mathrm{x}=5\) is:
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Let \(f(x)\) be a continuous function on \([a, b]\) and differentiable on \((a, b)\). Then, this function \(f(x)\) is strictly increasing in \((a, b)\) if
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The function \(\mathrm{f}(x)={x}^{2}-4x+6\) is increasing in the interval:
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Derivative of \(\cos \left(\sqrt{x}\right)\) is :
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If the local maximum value of the function \(f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}, x \in\left(0, \frac{\pi}{2}\right)\) is \(\frac{k}{e}\), then \(\left(\frac{k}{e}\right)^8+\frac{k^8}{e^5}+k^8\) is equal to
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If Rolle’s theorem holds for the function \(f(x) = \ x^{3} - ax^{2} + bx - 4,\) \(x\in \lbrack 1,2\rbrack\) with \(f'(\frac{4}{3}) = 0,\) then ordered pair \((a, b)\) is equal to
[JEE Main 2021, 25 Feb (Shift 1)]
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