Generating...                               s20quiz1315_n7

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

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

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

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

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

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

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

    $f'(x) =\displaystyle {{1}\over{5\,x^2-x+4}} $ $f'(x) =\displaystyle \left(10\,x-1\right)\,\ln \left\vert 5\,x^2-x+4\right\vert $ $f'(x) =\displaystyle 10\,x-1 $ $f'(x) =\displaystyle {{10\,x-1}\over{5\,x^2-x+4}} $

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

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

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

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

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

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

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

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


Department of Mathematics
Last modified: 2025-06-19