Generating...                               s20quiz1315_n9

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

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

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

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

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

  4. Find the derivative $f'(x)$ for $f(x) =\displaystyle \ln \left\vert 2\,x^2-x+1\right\vert $ .

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

  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( {{8\,x^3}\over{x^4+1}}+{{3\,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( {{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( -{{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) $

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

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

  8. 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} = -{{4\,y^3}\over{4\,x^3-1}} $ $\displaystyle \frac{dy}{dx} = -{{x^3}\over{y^3}} $ $\displaystyle \frac{dy}{dx} = -{{y^3}\over{x^3}} $ $\displaystyle \frac{dy}{dx} = y^4+x^4 $

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

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

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

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


Department of Mathematics
Last modified: 2026-02-06