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This is one of our many quizzes on Differentiation rules. Click on this to access more such competitive quizzes on Differentiation.

Differentiation rules Quiz 2

mathematics

 If $f''\left( x \right)=2{{\sec }^{2}}\left( x \right)\tan \left( x \right)$ then find the values of $f'\left( x \right)$and$f\left( x \right)$.

$f\left( x \right)=\tan \left( x \right) \\  f'(x)=se{{c}^{2}}(x) \\  f''(x)=2se{{c}^{2}}(x)tan(x) $

  If $y=2{{\cos }^{2}}\left( 3{{x}^{3}} \right)$then $\dfrac{dy}{dx}=?$

 Use differentiation basic rules and formulas.

$  y=2{{\cos }^{2}}\left( 3{{x}^{3}} \right) \\ \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( 2{{\cos }^{2}}\left( 3{{x}^{3}} \right) \right) \\ \Rightarrow 6\cos \left( 3{{x}^{3}} \right)\dfrac{d}{dx}\cos \left( 3{{x}^{3}} \right) \\ \Rightarrow -6\cos \left( 3{{x}^{3}} \right)\sin \left( 3{{x}^{3}} \right)9{{x}^{2}} \\ \Rightarrow -54{{x}^{2}}\cos \left( 3{{x}^{3}} \right)\sin \left( 3{{x}^{3}} \right) $

 if $f\left( x \right)={{x}^{\alpha }}$then, $f'\left( 2 \right)=?$

$f\left( x \right)={{x}^{\alpha }} \\  f'\left( x \right)=\alpha {{x}^{\alpha -1}} \\ $ Putting $x=2$ $ \\ f'\left( x \right)=\alpha {{\left( 2 \right)}^{\alpha -1}}$

 find derivative of $f\left( x \right)\text{ }=\text{ }\dfrac{\text{3}{{\text{x}}^{2}}}{x+25}$

 $f'\left( x \right)\text{ }=\text{ }\dfrac{g'\left( x \right)h\left( x \right)-\text{ }g\left( x \right)h'\left( x \right)\text{ }}{h{{\left( x \right)}^{2}}} \\  f'\left( x \right)=\dfrac{6x(x+25)-3{{x}^{2}}(1)}{{{(x+25)}^{2}}} \\  f'\left( x \right)=\dfrac{6{{x}^{2}}+150x-3{{x}^{2}}}{{{(x+25)}^{2}}} \\  f'\left( x \right)=\dfrac{3{{x}^{2}}+150x}{{{(x+25)}^{2}}} $ 

 Find the third derivative of the following function: $ \\ f(x)={{(2x-5)}^{3}}$ 

  $  f(x)={{(2x-5)}^{3}} \\  f'(x)=3{{(2x-54)}^{2}}.\dfrac{d}{dx}(2x-54)=3{{(2x-54)}^{2}}.2=6{{(2x-54)}^{2}} \\  f''(x)=\dfrac{d}{dx}\left( 6{{(2x-54)}^{2}} \right)=24(2x-54) \\  f'''(x)=\dfrac{d}{dx}24(2x-54)=\dfrac{d}{dx}\left( 48x-1296 \right)=48 $ 