Generating...                               s20quiz1315_n20

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

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

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

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

  3. If $\displaystyle y^4+x^4=1 $ , then find $\displaystyle \frac{dy}{dx}$ by implicit differentiation.

    $\displaystyle \frac{dy}{dx} = y^4+x^4 $ $\displaystyle \frac{dy}{dx} = -{{4\,y^3}\over{4\,x^3-1}} $ $\displaystyle \frac{dy}{dx} = -{{4\,x^3-1}\over{4\,y^3}} $ $\displaystyle \frac{dy}{dx} = -{{y^3}\over{x^3}} $ $\displaystyle \frac{dy}{dx} = 4\,x^3 $ $\displaystyle \frac{dy}{dx} = -{{x^3}\over{y^3}} $

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

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

  5. Find the derivative $f'(x)$ for $f(x) =\displaystyle \arctan e^ {- 3\,x } $ .

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

  6. If $\displaystyle y^2+x^2=x $ , then find $\displaystyle \frac{dy}{dx}$ by implicit differentiation.

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

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

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

  8. Find the derivative $f'(x)$ for $f(x) =\displaystyle \ln \cosh x $ .

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

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

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

  10. Find the derivative $f'(x)$ for $f(x) =\displaystyle \arctan \sqrt{x} $ .

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


Department of Mathematics
Last modified: 2026-05-20