Generating...                               s20quiz1315_n13

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

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

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

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

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

  4. 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 {{1}\over{e^{x}-e^ {- x }}} $ $f'(x) =\displaystyle \tanh x $ $f'(x) =\displaystyle \coth x $ $f'(x) =\displaystyle {{1}\over{e^{x}+e^ {- x }}} $

  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 x^{x} $ $f'(x) =\displaystyle \left(\ln x\right)^{x-1} $ $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. If $\displaystyle x^3\,y^3+\sinh x=y $ , then find $\displaystyle \frac{dy}{dx}$ by implicit differentiation.

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

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

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

  8. Let $\displaystyle y={{\left(x^2+4\right)^4\,\left(x^4+1\right)^3}\over{\sqrt{x-4}}} $ . Find the derivative in the form $y' = y (\cdots)$ by logarithmic differentiation.

    $\displaystyle y' = y \left( {{12\,x^3}\over{x^4+1}}+{{8\,x}\over{x^2+4}}-{{1}\over{2\,\left(x-4 \right)}} \right) $ $\displaystyle y' = y \left( {{12\,x^3}\over{x^4+1}}+{{8\,x}\over{x^2+4}}+{{1}\over{x-4}} \right) $ $\displaystyle y' = y \left( {{12\,x^3}\over{x^4+1}}+{{2\,x}\over{x^2+4}}-{{1}\over{2\,\left(x-4 \right)}} \right) $ $\displaystyle y' = y \left( {{4\,x^3}\over{x^4+1}}+{{8\,x}\over{x^2+4}}-{{1}\over{2\,\left(x-4 \right)}} \right) $ $\displaystyle y' = y \left( 12\,x^3\,\left(x^4+1\right)^2+8\,x\,\left(x^2+4\right)^3-{{1}\over{ 2\,\left(x-4\right)^{{{3}\over{2}}}}} \right) $ $\displaystyle y' = y \left( {{4\,x^3}\over{x^4+1}}+{{2\,x}\over{x^2+4}}+{{1}\over{x-4}} \right) $

  9. If $\displaystyle y^2+x^2=1 $ , then find $\displaystyle \frac{dy}{dx}$ by implicit differentiation.

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

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


Department of Mathematics
Last modified: 2025-12-16