Generating...                               s20quiz1315_n13

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

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

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

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

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

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

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

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

  5. 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-1} $ $f'(x) =\displaystyle x^{x} $ $f'(x) =\displaystyle \left(\ln x\right)^{x}\,\ln \ln x $ $f'(x) =\displaystyle \left(\ln x\right)^{x}\,\ln \ln x+\left(\ln x\right)^{x-1} $

  6. Find the derivative $f'(x)$ for $f(x) =\displaystyle \ln \sinh 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 {{e^{x}-e^ {- x }}\over{e^{x}+e^ {- x }}} $ $f'(x) =\displaystyle {{1}\over{e^{x}+e^ {- x }}} $

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

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

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

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

  9. Find the derivative $f'(x)$ for $f(x) =\displaystyle x^{\cosh 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 x^{\cosh x-1}\,\cosh x $ $f'(x) =\displaystyle {{\cosh x\,\left(\ln x\right)^{\cosh x-1}}\over{x}} $ $f'(x) =\displaystyle x^{\cosh x}\,\ln x\,\sinh x+x^{\cosh x-1}\,\cosh x $ $f'(x) =\displaystyle \left(\ln x\right)^{\cosh x}\,\sinh x\,\ln \ln x $

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


Department of Mathematics
Last modified: 2026-03-24