Differentiation

If $f(x)=\sum_{p=1}^7 p^2 \sin ^{-1}\left(\frac{4}{5} \sin (p x)-\frac{3}{5} \cos (p x)\right)$, then the value of $\frac{d f}{d x}$ at $x=1$ is [given that $\sin ^{-1}(\sin x)=x$ ])

If $x^2+y^2=t-\frac{1}{t}, x^4+y^4=t^2+\frac{1}{t^2}$, then $\frac{d y}{d x}=$

If $x=a(\cos \theta+\theta \sin \theta), y=f(\theta), f(2 \pi)=0$, $\frac{d y}{d x}=\frac{\tan \theta}{\theta}, \theta \neq 0$ and $\theta \neq(2 n+1) \frac{\pi}{2}$, then $f\left(\frac{\pi}{3}\right)=$

If $x \cos (k+y)=\cos y$, then $\frac{d y}{d x}$ at $y=\frac{\pi}{2}$ is

Let $f(x)=\sin x, g(x)=\cos x, h(x)=x^2$, then $\lim _{x \rightarrow 1} \frac{f(g(h(x)))-f(g(h(1)))}{x-1}=$

If $\frac{3 x+5}{(x+1)\left(2 x^2+3\right)}=\frac{A}{x+1}+\frac{B x+C}{2 x^2+3}$ and $f(x)=A x^3+B x^2+7 x+C$, then $5 C-f^{\prime}(-2)=$

\(\frac{d}{d x}\left[\left(x^{\frac{5}{2}}-x^{\frac{3}{2}}+1\right)\left(x^2-3 x+5\right)\right]=\)

If $\frac{d}{d x}\left(\frac{2 x+1}{(x+1)^2(x-2)}\right)=\frac{A}{(x-2)^2}+\frac{B}{(x+1)^3}+\frac{C}{(x+1)^2}$, then $A+B+C=$

If $f(x)=\frac{1+\sec x}{2(\sec x-1)}$ for $0

The value of $\frac{d}{d x}\left[\log \left(\sin \sqrt{\frac{x^2+1}{x^2+2}}\right)\right]$ when $x=\sqrt{2}$, is

If $f(x)=\log _e\left(e^{2 x}\left(\frac{3 x+5}{5-3 x}\right)^{2 / 3}\right), x \neq \frac{-5}{3}, \frac{5}{3}$, then the value of $\frac{d f}{d x}$ at $x=1$, is

If $x=\operatorname{cosec} \theta-\sin \theta, y=\operatorname{cosec}^{2022} \theta-\sin ^{2022} \theta$ and $\left(\frac{d y}{d x}\right)^2=\frac{k\left(y^2+4\right)}{g(x)}$ where $k \in R$, then $10+k-g(2022)=$

If $a f(x)+b f\left(\frac{1}{x}\right)=x+1$, and $\frac{d}{d x}\left(x^2 f(x)\right)=2 x^2+2 x+\frac{1}{3}$, then $a-b$

If $f(x)=\sin \left(\cosh \left(\frac{x^2+1}{x^2+2}\right)\right)$, then $f^{\prime}(1)=$

If $y(x)=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$ and $\left(1+a^2 x^2\right) y^{\prime \prime}+g(x) y^{\prime}=0$ then, the sum of the roots of the equation $1+a^2 x^2+g(x)=0$ is

If $\operatorname{Lt}_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=e^x(x+1)$ and $f(0)=0$, then $\frac{d}{d x}\left(f(x) e^{-x}\right)+\frac{d}{d x}\left(\frac{f(x)}{x}\right)=$

Let $f: R \rightarrow R$ be defined by $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all $x$ and $y$. If $f^{\prime}(0)$ exists and equals -1 and $f(0)=1$, then $f(2)=$

\(\frac{d}{d x}\left[\operatorname{cosech}^{-1}(\tan 2 x)\right]=\)

If $2^x+2^y=2^{x+y}$, then $\frac{d y}{d x}=$

If $p(x)$ be a polynomial satisfying $p(2 x)=p^{\prime}(x) \cdot p^{\prime \prime}(x)$, then $\sum_{x=1}^5 p(x)=$

If $x \sqrt{1+y}+y \sqrt{1+x}=0$, then $\frac{d y}{d x}=$

$$ \begin{aligned} & \text { If }\left(\frac{d y}{d x}\right)=\frac{1}{\left(\frac{d x}{d y}\right)} \text { and } \frac{d^2 x}{d y^2}\left(\frac{d y}{d x}\right)^3+\frac{d^2 y}{d x^2}=k \text {, then } \\ & e^{k f(x)}-k f(x)= \end{aligned}...

If $f(x)=\frac{x-1}{e^x}$, then $f^{\prime}(0)+f^{\prime \prime}(0)=$

Match the functions of List-I with derivates given in List-II
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Let $f(x)$ and $g(x)$ be twice differentiable functions such that $f(x)=x^2+g^{\prime}(1) x+g^{\prime \prime}(2)$ and $g(x)=f(1) x^2+x f^{\prime}(x)+f^{\prime \prime}(x)$. Then $f(x)-g(x)=$

If $y=\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\log \sqrt{1-x^2}$, then $\frac{d y}{d x}=$

let $g(x) \neq 0, g^{\prime}(x) \neq 0, f(x) \neq 0, f^{\prime}(x) \neq 0$. If
$F(x)=f(x) g(x), G(x)=f^{\prime}(x) g^{\prime}(x)$ and
$F^{\prime}(x)=G(x) H(x)=F(x) K(x)$, then $H(x)+K(x)=$

$$ \begin{aligned} & \text { If } f(x)=\tan ^{-1}\left(\frac{1}{\sin ^2 x+\sin x+1}\right) \\ & \quad+\tan ^{-1}\left(\frac{1}{\sin ^2 x+3 \sin x+3}\right)+\tan ^{-1} \end{aligned} $$
$\left(\frac{1}{\sin ^2 x+5 \sin x+7}\right)+\ldots+$ up...

If $\alpha$ is such a minimum value for which the inverse of $f(x)=x^2+3 x-3$ exists in $[\alpha, \infty)$ and $g$ is the inverse of the $f$, then at $x=\alpha+\frac{5}{2}, \frac{d g}{d x}$