Generating...                               s20quiz1315_n3

  1. 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 }}} $

  2. Find the derivative $f'(x)$ for $f(x) =\displaystyle \ln \left\vert 4\,x^2-2\,x+5\right\vert $ .

    $f'(x) =\displaystyle {{1}\over{4\,x^2-2\,x+5}} $ $f'(x) =\displaystyle \left(8\,x-2\right)\,\ln \left\vert 4\,x^2-2\,x+5\right\vert $ $f'(x) =\displaystyle 8\,x-2 $ $f'(x) =\displaystyle {{8\,x-2}\over{4\,x^2-2\,x+5}} $

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

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

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

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

    $\displaystyle y' = y \left( {{4\,x^3}\over{x^4+1}}-{{9\,x^2}\over{x^3+4}}+{{6\,x}\over{x^2-2}} \right) $ $\displaystyle y' = y \left( {{4\,x^3}\over{x^4+1}}+{{3\,x^2}\over{x^3+4}}+{{2\,x}\over{x^2-2}} \right) $ $\displaystyle y' = y \left( {{8\,x^3}\over{x^4+1}}-{{9\,x^2}\over{x^3+4}}+{{6\,x}\over{x^2-2}} \right) $ $\displaystyle y' = y \left( {{8\,x^3}\over{x^4+1}}-{{9\,x^2}\over{x^3+4}}+{{2\,x}\over{x^2-2}} \right) $ $\displaystyle y' = y \left( -{{9\,x^2}\over{\left(x^3+4\right)^4}}+6\,x\,\left(x^2-2\right)^2+8 \,x^3\,\left(x^4+1\right) \right) $ $\displaystyle y' = y \left( {{8\,x^3}\over{x^4+1}}+{{3\,x^2}\over{x^3+4}}+{{6\,x}\over{x^2-2}} \right) $

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

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

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

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

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

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

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

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


Department of Mathematics
Last modified: 2026-03-24