Generating...                               s20quiz1315_n14

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

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

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

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

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

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

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

  5. 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 {{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 }}} $

  6. Find the derivative $f'(x)$ for $f(x) =\displaystyle \ln \left\vert 3\,x^2+4\,x-3\right\vert $ .

    $f'(x) =\displaystyle {{6\,x+4}\over{3\,x^2+4\,x-3}} $ $f'(x) =\displaystyle 6\,x+4 $ $f'(x) =\displaystyle {{1}\over{3\,x^2+4\,x-3}} $ $f'(x) =\displaystyle \left(6\,x+4\right)\,\ln \left\vert 3\,x^2+4\,x-3\right\vert $

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

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

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

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

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

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

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

    $\displaystyle y' = y \left( -{{9\,x^2}\over{x^3-3}}+{{x}\over{x^2-1}}+{{1}\over{x-3}} \right) $ $\displaystyle y' = y \left( {{3\,x^2}\over{x^3-3}}+{{2\,x}\over{x^2-1}}+{{1}\over{x-3}} \right) $ $\displaystyle y' = y \left( {{3\,x^2}\over{x^3-3}}+{{x}\over{x^2-1}}+{{4}\over{x-3}} \right) $ $\displaystyle y' = y \left( -{{9\,x^2}\over{\left(x^3-3\right)^4}}+{{x}\over{\sqrt{x^2-1}}}+4\, \left(x-3\right)^3 \right) $ $\displaystyle y' = y \left( -{{9\,x^2}\over{x^3-3}}+{{2\,x}\over{x^2-1}}+{{4}\over{x-3}} \right) $ $\displaystyle y' = y \left( -{{9\,x^2}\over{x^3-3}}+{{x}\over{x^2-1}}+{{4}\over{x-3}} \right) $


Department of Mathematics
Last modified: 2026-07-16