লঘুমান গুরুমান বিষয়ক
y=lnx+xy=\ln{x}+x y=lnx+x হলে, xdydx−y= ?x\frac{dy}{dx}-y=\ ?xdxdy−y= ?
1+lnx1+\ln{x} 1+lnx
lnx\\\ln{x} lnx
1−lnx\\1-\ln{x} 1−lnx
y\\y y
Given,
y=lnx+x
dydx=1x+1∴xdydx−y=x(1x+1)−lnx−x=1−lnx\\\frac{dy}{dx}=\frac{1}{x}+1 \\\therefore x\frac{dy}{dx}-y=x\left(\frac{1}{x}+1\right)-\ln{x}-x=1-\ln{x} dxdy=x1+1∴xdxdy−y=x(x1+1)−lnx−x=1−lnx
দৃশ্যকল্প-I: y(x+1)(x+2)−x+4 y(x+1)(x+2)-x+4 y(x+1)(x+2)−x+4
দৃশ্যকল্প-II: g(x)=3x3−6x2−5x+1 \mathrm{g}(\mathrm{x})=3 \mathrm{x}^{3}-6 \mathrm{x}^{2}-5 \mathrm{x}+1 g(x)=3x3−6x2−5x+1
Let f(x)={x3/5x≤1−(x−2)3x>1f\left( x \right) =\left\{ \begin{matrix} { x }^{ { 3 }/{ 5 } }\quad \quad \quad x\le 1 \\ -{ \left( x-2 \right) }^{ 3 }\quad x>1 \end{matrix} \right. f(x)={x3/5x≤1−(x−2)3x>1
then the number of critical points on the graph of the function is
If for all x,yx, yx,y the function f is defined by; f(x)+f(y)+f(x)⋅f(y)=1f(x)+f(y)+f(x)\cdot f(y)=1f(x)+f(y)+f(x)⋅f(y)=1 and f(x)>0f(x) > 0f(x)>0.When f(x)f(x)f(x) is differentiable f′(x)=f'(x)= f′(x)=,
x এর কোন মানের জন্য f(x)=xlnx f{\left ( x \right )} = \frac{x}{\ln{x}} f(x)=lnxx সর্বনিম্ন হবে-