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This is one of our many quizzes on Sum and difference rule. Click on this to access more such competitive quizzes on Calculus.

Sum and difference rule Quiz 2

mathematics

 What is the derivative of $\dfrac{d}{dx}\left[ f(x)\pm g(-x) \right]$ using sum and difference rules?

 Use chain rule to differentiate the function like f(ax) where ‘a’ belongs to any real number.We should also know that, $\left( \pm \right)\times \left( - \right)=\mp $ to solve this question.

  $ \dfrac{d}{dx}\left[ f(x)\pm g(-x) \right]  \\ \Rightarrow \dfrac{d[f(x)]}{dx}\pm \dfrac{d\left[ g(-x) \right]}{dx}  \\ \Rightarrow \dfrac{d[f(x)]}{dx}\pm \left( -\dfrac{d\left[ g(x) \right]}{dx} \right)  \\ \Rightarrow \dfrac{d[f(x)]}{dx}\mp \dfrac{d\left[ g(x) \right]}{dx} $

 Find the derivative of ${{x}^{3}}+2\sqrt{2}{{x}^{\frac{4}{3}}}-\dfrac{3x}{\sqrt{5}}$ .

 We need to know one of the standard derivative: $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$. Using sum rule, first we must separate them into individual terms then apply the above rule to each of them. 

  $ \dfrac{d\left( {{x}^{3}}+2\sqrt{2}{{x}^{\frac{4}{3}}}-\dfrac{3x}{\sqrt{5}} \right)}{dx} \\ \dfrac{d\left( {{x}^{3}} \right)}{dx}+\dfrac{d\left( 2\sqrt{2}{{x}^{\frac{4}{3}}} \right)}{dx}+\dfrac{d\left( -\dfrac{3x}{\sqrt{5}} \right)}{dx} \\ 3{{x}^{2}}+2\sqrt{2}\dfrac{d\left( {{x}^{\frac{4}{3}}} \right)}{dx}-\dfrac{3}{\sqrt{5}}\dfrac{d\left( x \right)}{dx} \\ 3{{x}^{2}}+2\sqrt{2}\times \dfrac{4}{3}\times {{x}^{\frac{1}{3}}}-\dfrac{3}{\sqrt{5}} \\ 3{{x}^{2}}+\dfrac{8\sqrt{2}}{3}\times {{x}^{\frac{1}{3}}}-\dfrac{3}{\sqrt{5}} $

 Find the sum of the derivatives of ${{x}^{3}}+2{{x}^{2}}-\dfrac{x\sqrt{x}}{3}$ and ${{x}^{4}}+2{{x}^{3}}-\dfrac{{{x}^{\frac{4}{3}}}}{2}$.

 We need to know one of the standard derivative: $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$. Using sum rule, first we would separate them into individual term then apply the above rule to each of them. At last, sum both the derivative to get the answer.

 $ d\dfrac{\left( {{x}^{3}}+2{{x}^{2}}-\dfrac{x\sqrt{x}}{3} \right)}{dx}  \\ \Rightarrow \dfrac{d\left( {{x}^{3}} \right)}{dx}+\dfrac{d\left( 2{{x}^{2}} \right)}{dx}+\dfrac{d\left( -\dfrac{x\sqrt{x}}{3} \right)}{dx}  \\ \Rightarrow 3{{x}^{2}}+2\dfrac{d\left( {{x}^{2}} \right)}{dx}-\dfrac{1}{3}\dfrac{d\left( {{x}^{\frac{3}{2}}} \right)}{dx}  \\ \Rightarrow 3{{x}^{2}}+4x-\dfrac{1}{3}\times \dfrac{3}{2}\times {{x}^{\frac{1}{2}}}  \\ \Rightarrow 3{{x}^{2}}+4x-\dfrac{1}{2}\times \sqrt{x} \\ \dfrac{d\left( {{x}^{4}}+2{{x}^{3}}-\dfrac{{{x}^{\frac{4}{3}}}}{2} \right)}{dx}  \\ \Rightarrow \dfrac{d\left( {{x}^{4}} \right)}{dx}+\dfrac{d\left( 2{{x}^{3}} \right)}{dx}+\dfrac{d\left( -\dfrac{{{x}^{\frac{4}{3}}}}{2} \right)}{dx}  \\ \Rightarrow 4{{x}^{3}}+6{{x}^{2}}-\dfrac{1}{2}\times \dfrac{4}{3}\times {{x}^{\frac{1}{3}}}  \\ \Rightarrow 4{{x}^{3}}+6{{x}^{2}}-\dfrac{2}{3}\times {{x}^{\frac{1}{3}}} \\ \text{Sum}= 3{{x}^{2}}+4x-\dfrac{1}{2}\times \sqrt{x}+4{{x}^{3}}+6{{x}^{2}}-\dfrac{2}{3}\times {{x}^{\frac{1}{3}}}  \\ \Rightarrow 4{{x}^{3}}+9{{x}^{2}}+4x-\dfrac{\sqrt{x}}{2}-\dfrac{2\sqrt[3]{x}}{3} $

 Find the derivative of $f(x)=4{{x}^{5}}-{{\sin }^{2}}\sqrt{x}+\dfrac{\sin 2x+{{\cos }^{2}}3x}{3}$

 We need to know one of the standard derivative: $1.\text{ }\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}\text{ 2}\text{. }\dfrac{d\left( \sin \text{ x} \right)}{dx}=\cos \text{ x 3}\text{. }\dfrac{d\left( \cos \text{ x} \right)}{dx}=-\sin \text{ x}$. Using sum rule, first we would separate them into individual terms then apply the above rule to each of them.

 $ \dfrac{d\left( 4{{x}^{5}}-{{\sin }^{2}}\sqrt{x}+\dfrac{\sin 2x+{{\cos }^{2}}3x}{3} \right)}{dx}  \\ \Rightarrow \dfrac{d\left( 4{{x}^{5}}-{{\sin }^{2}}\sqrt{x}+\dfrac{\sin 2x}{3}+\dfrac{{{\cos }^{2}}3x}{3} \right)}{dx}  \\ \Rightarrow \dfrac{d\left( 4{{x}^{5}} \right)}{dx}+\dfrac{d\left( -{{\sin }^{2}}\sqrt{x} \right)}{dx}+\dfrac{d\left( \dfrac{\sin 2x}{3} \right)}{dx}+\dfrac{d\left( \dfrac{{{\cos }^{2}}3x}{3} \right)}{dx}  \\ \Rightarrow 4\times 5{{x}^{4}}-2\sin \sqrt{x}\times \text{cos }\sqrt{x}\times \dfrac{1}{2}{{\left( x \right)}^{-\frac{1}{2}}}+\dfrac{1}{3}\cos \text{ 2x}\times \text{2+}\dfrac{1}{3}\times 2\cos \text{ }3x\times (-\sin \text{ 3x)}\times \text{3} \\ \text{=20}{{\text{x}}^{4}}-\dfrac{\sin \text{ 2}\sqrt{x}}{2\sqrt{x}}+\dfrac{2}{3}\cos \text{ 2x-sin 6x} $

 Find the derivative of $f(x)={{\sin }^{2}}x\cos x+{{\tan }^{\frac{1}{2}}}x-\cos \sqrt{x}+ \\ \dfrac{\sin 2x}{\cos x}$

 We need to know one of the standard derivatives: $1.\text{ }\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}\text{ 2}\text{. }\dfrac{d\left( \sin \text{ x} \right)}{dx}=\cos \text{ x 3}\text{. }\dfrac{d\left( \cos \text{ x} \right)}{dx}=-\sin \text{ x 4}\text{. }\dfrac{d\left( \tan \text{ x} \right)}{dx}={{\sec }^{2}}\text{ x}$. Using sum rule, first we would separate them into individual term then apply the above rule to each of them.

$ \dfrac{d\left( {{\sin }^{2}}x\cos x+{{\tan }^{\frac{1}{2}}}x-\cos \sqrt{x}+\dfrac{\sin 2x}{\cos x} \right)}{dx}  \\ \Rightarrow \dfrac{d\left( {{\sin }^{2}}x\cos x \right)}{dx}+\dfrac{d\left( {{\tan }^{\frac{1}{2}}}x \right)}{dx}+\dfrac{d\left( -\cos \sqrt{x} \right)}{dx}+\dfrac{d\left( \dfrac{\sin 2x}{\cos x} \right)}{dx}  \\ \Rightarrow \dfrac{d\left( {{\sin }^{2}}x\cos x \right)}{dx}+\dfrac{d\left( {{\tan }^{\frac{1}{2}}}x \right)}{dx}+\dfrac{d\left( -\cos \sqrt{x} \right)}{dx}+\dfrac{d\left( \dfrac{2\sin x\text{ cosx}}{\cos x} \right)}{dx}  \\ \Rightarrow \left( \cos \text{ x} \right)\times 2\sin \text{ x cos x+si}{{\text{n}}^{2}}\text{ x (-sin x)+}\dfrac{1}{2}\left( {{\tan }^{-\frac{1}{2}}}\text{ x} \right)\text{ se}{{\text{c}}^{2}}\text{ x-}\left( \text{-sin }\sqrt{x} \right)\times \dfrac{1}{2}\times {{x}^{-\frac{1}{2}}}+2\cos \text{ x}  \\ \Rightarrow \sin \text{ 2x cos x-si}{{\text{n}}^{3}}\text{ x+}\dfrac{\text{se}{{\text{c}}^{2}}\text{ x}}{2\sqrt{\tan \text{ x}}}+\dfrac{\sin \text{ }\sqrt{x}}{2\sqrt{x}}+2\cos \text{ x} $<br>