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

    $\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}} $ $\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} = \cos x\,\cos y $ $\displaystyle \frac{dy}{dx} = \sin x\,\cos y $

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

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

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

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

  4. 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 {{\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 $ $f'(x) =\displaystyle 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) $

  5. 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-1} $ $f'(x) =\displaystyle \left(\ln x\right)^{x}\,\ln \ln x $ $f'(x) =\displaystyle x^{x}\,\ln x+x^{x} $ $f'(x) =\displaystyle x^{x} $

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

    $f'(x) =\displaystyle {{2\,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 {{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}} $

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

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

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

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

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

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

  10. Find the derivative $f'(x)$ for $f(x) =\displaystyle \ln \left\vert 5\,x^2-3\,x+2\right\vert $ .

    $f'(x) =\displaystyle 10\,x-3 $ $f'(x) =\displaystyle {{1}\over{5\,x^2-3\,x+2}} $ $f'(x) =\displaystyle {{10\,x-3}\over{5\,x^2-3\,x+2}} $ $f'(x) =\displaystyle \left(10\,x-3\right)\,\ln \left\vert 5\,x^2-3\,x+2\right\vert $


Department of Mathematics
Last modified: 2026-08-24