Generating...                               s20quiz1315_n2

  1. If $\displaystyle \sinh y+x^3\,y^3=y $ , then find $\displaystyle \frac{dy}{dx}$ by implicit differentiation.

    $\displaystyle \frac{dy}{dx} = -{{\cosh y+3\,x^3\,y^2}\over{3\,x^2\,y^3-1}} $ $\displaystyle \frac{dy}{dx} = -{{\cosh y+3\,x^3\,y^2-1}\over{3\,x^2\,y^3}} $ $\displaystyle \frac{dy}{dx} = \sinh y+x^3\,y^3 $ $\displaystyle \frac{dy}{dx} = 3\,x^2\,y^3 $ $\displaystyle \frac{dy}{dx} = -{{3\,x^2\,y^3}\over{\cosh y+3\,x^3\,y^2-1}} $ $\displaystyle \frac{dy}{dx} = -{{3\,x^2\,y^3-1}\over{\cosh y+3\,x^3\,y^2}} $

  2. Find the derivative $f'(x)$ for $f(x) =\displaystyle \arctan x^2 $ .

    $f'(x) =\displaystyle {{1}\over{x^4+1}} $ $f'(x) =\displaystyle {{x^2}\over{x^4+1}} $ $f'(x) =\displaystyle {{2\,x}\over{\tan x^2}} $ $f'(x) =\displaystyle -{{2\,x\,\left(\sec x^2\right)^2}\over{\tan ^2x^2}} $ $f'(x) =\displaystyle {{2\,x}\over{x^4+1}} $ $f'(x) =\displaystyle {{x^2}\over{\tan x^2}} $

  3. Find the derivative $f'(x)$ for $f(x) =\displaystyle x^{\cosh x} $ .

    $f'(x) =\displaystyle \left(\ln x\right)^{\cosh x}\,\sinh x\,\ln \ln x $ $f'(x) =\displaystyle \left(\ln x\right)^{\cosh x}\,\left(\sinh x\,\ln \ln x+{{\cosh x }\over{x\,\ln x}}\right) $ $f'(x) =\displaystyle {{\cosh x\,\left(\ln x\right)^{\cosh x-1}}\over{x}} $ $f'(x) =\displaystyle x^{\cosh x-1}\,\cosh x $ $f'(x) =\displaystyle x^{\cosh x}\,\ln x\,\sinh x+x^{\cosh x-1}\,\cosh x $

  4. If $\displaystyle \sqrt{y}+\sqrt{x}=x $ , then find $\displaystyle \frac{dy}{dx}$ by implicit differentiation.

    $\displaystyle \frac{dy}{dx} = {{1}\over{2\,\sqrt{x}}} $ $\displaystyle \frac{dy}{dx} = {{\left(2\,\sqrt{x}-1\right)\,\sqrt{y}}\over{\sqrt{x}}} $ $\displaystyle \frac{dy}{dx} = {{\sqrt{x}}\over{\left(2\,\sqrt{x}-1\right)\,\sqrt{y}}} $ $\displaystyle \frac{dy}{dx} = \sqrt{y}+\sqrt{x} $ $\displaystyle \frac{dy}{dx} = {{\sqrt{y}}\over{2\,\sqrt{x}\,\sqrt{y}-\sqrt{x}}} $ $\displaystyle \frac{dy}{dx} = {{2\,\sqrt{x}\,\sqrt{y}-\sqrt{x}}\over{\sqrt{y}}} $

  5. Find the derivative $f'(x)$ for $f(x) =\displaystyle x^{x} $ .

    $f'(x) =\displaystyle \left(\ln x\right)^{x}\,\ln \ln x $ $f'(x) =\displaystyle x^{x}\,\ln x+x^{x} $ $f'(x) =\displaystyle x^{x} $ $f'(x) =\displaystyle \left(\ln x\right)^{x}\,\left(\ln \ln x+{{1}\over{\ln x}} \right) $ $f'(x) =\displaystyle \left(\ln x\right)^{x-1} $

  6. Find the derivative $f'(x)$ for $f(x) =\displaystyle \left(\ln x\right)^{x} $ .

    $f'(x) =\displaystyle x^{x}\,\left(\ln x+1\right) $ $f'(x) =\displaystyle \left(\ln x\right)^{x}\,\ln \ln x $ $f'(x) =\displaystyle \left(\ln x\right)^{x-1} $ $f'(x) =\displaystyle \left(\ln x\right)^{x}\,\ln \ln x+\left(\ln x\right)^{x-1} $ $f'(x) =\displaystyle x^{x} $

  7. Find the derivative $f'(x)$ for $f(x) =\displaystyle \arcsin e^{2\,x} $ .

    $f'(x) =\displaystyle {{1}\over{\sqrt{1-e^{4\,x}}}} $ $f'(x) =\displaystyle {{2\,e^{2\,x}}\over{\sin e^{2\,x}}} $ $f'(x) =\displaystyle {{e^{2\,x}}\over{\sin e^{2\,x}}} $ $f'(x) =\displaystyle -{{2\,e^{2\,x}\,\cos e^{2\,x}}\over{\sin ^2e^{2\,x}}} $ $f'(x) =\displaystyle {{e^{2\,x}}\over{\sqrt{1-e^{4\,x}}}} $ $f'(x) =\displaystyle {{2\,e^{2\,x}}\over{\sqrt{1-e^{4\,x}}}} $

  8. If $\displaystyle {{1}\over{y}}+{{1}\over{x}}=1 $ , then find $\displaystyle \frac{dy}{dx}$ by implicit differentiation.

    $\displaystyle \frac{dy}{dx} = -{{x^2}\over{y^2}} $ $\displaystyle \frac{dy}{dx} = -{{1}\over{x^2}} $ $\displaystyle \frac{dy}{dx} = {{1}\over{y}}+{{1}\over{x}} $ $\displaystyle \frac{dy}{dx} = -{{\left(x^2+1\right)\,y^2}\over{x^2}} $ $\displaystyle \frac{dy}{dx} = -{{y^2}\over{x^2}} $ $\displaystyle \frac{dy}{dx} = -{{x^2}\over{\left(x^2+1\right)\,y^2}} $

  9. Find the derivative $f'(x)$ for $f(x) =\displaystyle \ln \left\vert 3\,x^2+2\,x-3\right\vert $ .

    $f'(x) =\displaystyle 6\,x+2 $ $f'(x) =\displaystyle \left(6\,x+2\right)\,\ln \left\vert 3\,x^2+2\,x-3\right\vert $ $f'(x) =\displaystyle {{1}\over{3\,x^2+2\,x-3}} $ $f'(x) =\displaystyle {{6\,x+2}\over{3\,x^2+2\,x-3}} $

  10. Find the derivative $f'(x)$ for $f(x) =\displaystyle \ln \left(e^{x}+e^ {- x }\right) $ .

    $f'(x) =\displaystyle {{e^{x}+e^ {- x }}\over{e^{x}-e^ {- x }}} $ $f'(x) =\displaystyle \coth x $ $f'(x) =\displaystyle \tanh x $ $f'(x) =\displaystyle {{1}\over{e^{x}-e^ {- x }}} $ $f'(x) =\displaystyle {{1}\over{e^{x}+e^ {- x }}} $


Department of Mathematics
Last modified: 2025-05-04