14-Application of Derivatives

Application of Derivatives

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How to prepare for Application of Derivatives for JEE Mains
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Need help in Question
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How to prove it ?
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pvashistha
Book or resource to use for calculus
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raheesh6264
If an invertible function f(x) satisfies x + |f(x)| = 2f(x) then f-1(x) satisfies
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Amodini
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Please explain this question with proper reasoning and explanation
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Sneha Duhan
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If $f(x) = x ^5/5 + x ^ 4/4 + x ^ 3 + k x^2/2 + x$ be a real valued function. Then the greatest value of $k^2$ for which f(x) is an increasing ....?
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Sneha Duhan
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Find the minimum and maximum value of ?
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Explain partial derivatives by taking an instance in day today life and where we should use it
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Amodini
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A rod of fixed length $k$ has its ends sliding along the co-ordinate axes in $1^{\text {st }}$ quadrant such that rod meet the $x$-axis at $A(a, 0)$
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deveshgarg2008
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Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function such that $f(1)=2$. $ 6 \int_1^x f(t) d t=3 x f(x)-x^3 $ for all $x \geq 1$,
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Minimizing Folded Area: Leaf Corner Fold Problem (Finding p+q when area is minimum)
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Explain Increasing/Decreasing Functions
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Find absolute maximum values of three related exponential functions on [0,1] and determine their relationship
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Rolle's Theorem - If $P(x)=51 x^{101}-2323 x^{100}-45 x+1035$, then Rolle's is applicable for
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Prove that a conical tent of a given capacity will required the least amount of canvas when the height is √2 times the radius of the base?
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A polynomial of degree 4 has absolute minimum of -123 at \( x=-4 \) and a relative minimum of 2 at \( x=1 \). The second derivative of the polynomial
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A sheet of area 102 is cut to form a closed box with a square base. What will be the maximum volume of the box?
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For what \(\theta\), \(x\) will be minimum: \[ x=\frac{-1+2 \cos ^{2} \theta}{\cos \theta} \]
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Arnav
$f(x)= \begin{cases}4 x-x^3+\ln \left(a^2-3 a+3\right), & 0 \leq x<3 \\ x-18, & x \geq 3\end{cases}$. Complete set of values of a such that $f(x)$ has
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_Vaishnavi
If the function $f(x)=\left(\frac{1}{x}\right)^{2 x} ; x>0$ attains the maximum value at $x=\frac{1}{\mathrm{e}}$ then :
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