ত্রিকোনমিতিক ফাংশনের অন্তরজ
ln(cosx)x\frac{\ln (\cos x)}{x}xln(cosx)
xxx এর সাপেক্ষে অন্তরক সহগ নিচের কোনটি?
{2xtanx+ln(sinx)x2 \frac{\{2x \tan x+\ln (\sin x)}{x^{2}}x2{2xtanx+ln(sinx)
{xtanx−ln(cosx)x4 \frac{\{x \tan x-\ln (\cos x)}{x^{4}}x4{xtanx−ln(cosx)
-{xtanx+ln(cosx)x2 \frac{\{x \tan x+\ln (\cos x)}{x^{2}}x2{xtanx+ln(cosx)
{xtanx−ln(cosx)x2 \frac{\{x \tan x-\ln (\cos x)}{x^{2}}x2{xtanx−ln(cosx)
Solve:
ln(cosx)xddx{ln(cosx)x}=xddx{ln(cosx)−ln(cosx)ddx(x)x2=x1cosx(−sinx)−ln(cosx)⋅1x2=−{xtanx+ln(cosx)x2 \begin{array}{l} \frac{\ln (\cos x)}{x} \\ \frac{d}{d x}\left\{\frac{\ln (\cos x)}{x}\right\} \\ =\frac{x \frac{d}{d x}\left\{\ln (\cos x)-\ln (\cos x) \frac{d}{d x}(x)\right.}{x^{2}} \\ =\frac{x \frac{1}{\cos x}(-\sin x)-\ln (\cos x) \cdot 1}{x^{2}} \\ =-\frac{\{x \tan x+\ln (\cos x)}{x^{2}} \\ \end{array} xln(cosx)dxd{xln(cosx)}=x2xdxd{ln(cosx)−ln(cosx)dxd(x)=x2xcosx1(−sinx)−ln(cosx)⋅1=−x2{xtanx+ln(cosx)
If the functions f(x)=sin(x+a) \displaystyle f\left ( x \right )=\sin \left ( x+a \right ) f(x)=sin(x+a) and g(x)=bsinx+ccosx \displaystyle g\left ( x \right )=b\sin x+c\cos x g(x)=bsinx+ccosx satisfy f(0)=g(0) \displaystyle f\left ( 0 \right )=g\left ( 0 \right ) f(0)=g(0) and f′(0)=g′(0) \displaystyle {f}'\left ( 0 \right )={g}'\left ( 0 \right ) f′(0)=g′(0) then
dydx\displaystyle\frac{dy}{dx}dxdy at t=π4\displaystyle t=\frac{\pi}{4}t=4π for x=a[cost+12logtan2t2]\displaystyle x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]x=a[cost+21logtan22t] and y=asinty=a\sin{t}y=asint is
y=ln(cosx) y=\ln (\cos x) y=ln(cosx) হলে, dydx \frac{d y}{d x} dxdy এর মান কত?
y=sin(1x) y = \sin{\left ( \frac{1}{x} \right )} y=sin(x1) হলে dydx \frac{dy}{dx} dxdy এর মান-
নিচের কোনটি সঠিক?