Generating...                               s20quiz1315_n10

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

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

  2. 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 }}} $

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

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

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

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

  5. Find the derivative $f'(x)$ for $f(x) =\displaystyle \arccos x^3 $ .

    $f'(x) =\displaystyle {{3\,x^2}\over{\cos x^3}} $ $f'(x) =\displaystyle -{{x^3}\over{\sqrt{1-x^6}}} $ $f'(x) =\displaystyle {{3\,x^2\,\sin x^3}\over{\cos ^2x^3}} $ $f'(x) =\displaystyle -{{3\,x^2}\over{\sqrt{1-x^6}}} $ $f'(x) =\displaystyle -{{1}\over{\sqrt{1-x^6}}} $ $f'(x) =\displaystyle {{x^3}\over{\cos x^3}} $

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

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

  7. Find the derivative $f'(x)$ for $f(x) =\displaystyle \arctan 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{\tan e^{2\,x}}} $ $f'(x) =\displaystyle {{e^{2\,x}}\over{e^{4\,x}+1}} $ $f'(x) =\displaystyle -{{2\,e^{2\,x}\,\left(\sec e^{2\,x}\right)^2}\over{\tan ^2e^{2\,x} }} $ $f'(x) =\displaystyle {{2\,e^{2\,x}}\over{e^{4\,x}+1}} $

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

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

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

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

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


Department of Mathematics
Last modified: 2025-09-14