G Tewani Solutions for Chapter: Methods of Differentiation, Exercise 1: DPP

The derivative of $\cos \left(2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right)-2{\cos }^{-1}\sqrt{\frac{1-x}{2}}$ w.r.t. $x$ is 

If $y=\frac{x^2}{2}+\frac{1}{2}x\sqrt{x^2+1}+\ln \sqrt{x+\sqrt{x^2+1}}$, then the value of $x{y}^{\text{'}}+\log y^{\text{'}}$ is

Let $g\left(x\right)=f\left(x\right)\sin x$, where $f\left(x\right)$ is a twice differentiable function on $\left(-\infty , \infty \right)$ such that $f^{\text{'}}\left(-\pi \right)=1$. Then the value of $g^{"}\left(-\pi \right)$ equals

If $f\left(x\right)={\log }_x\left({\log }_{e}x\right)$, then $f^{\text{'}}\left(x\right)$ at $x=e$ is:

Let $g\left(x\right)=e^{f\left(x\right)}$ and $f\left(x+1\right)=x+f\left(x\right) \forall x\in R$. If $n\in I^{+},$ then find the value of $\frac{g^{\text{'}}\left(n+\frac{1}{2}\right)}{g\left(n+\frac{1}{2}\right)}-\frac{g^{\text{'}}\left(\frac{1}{2}\right)}{g\left(\frac{1}{2}\right)}$.

Find the value of $\frac{d}{dx}\left[{\cos }^{-1}(x\sqrt{x}-\sqrt{(1-x)\left(1-x^2\right)})\right]$.

If $y={\cos }^{-1}\sqrt{\frac{\sqrt{1+x^2}+1}{2\sqrt{1+x^2}}}$, then $\frac{dy}{dx}$ is equal to:

Suppose that $f\left(x\right)$ is a differentiable invertible function such that $f^{\text{'}}\left(x\right)\ne 0$ and $h^{\text{'}}\left(x\right)=f\left(x\right).$ Given that $f\left(1\right)=f^{\text{'}}\left(1\right)=1, h\left(1\right)=0$ and $g\left(x\right)$ is the inverse of $f\left(x\right)$. Let $G\left(x\right)=x^{2}g\left(x\right)-xh\left(g\right(x\left)\right),\forall x\in R$. Which of the following is/are correct?