Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Second derivative test. Click on this to access more such competitive quizzes on Differential Calculus.

Second derivative test Quiz 2

mathematics

 What is the second derivative of  ${{x}^{2}}-4\sin x+2\cos x+{{e}^{x}}$

 As it is asked in the question to find the second derivative therefore differentiate the given function twice to get the second derivative.

  $f(x)={{x}^{2}}-4\sin x+2\cos x+{{e}^{x}} \\ \text{Differentiate wrt x:} \\ f'(x)=2x-4\cos x-2\sin x+{{e}^{x}} \\ \text{Differentiate wrt x:} \\ f''(x)=2+4\sin x-2\cos +{{e}^{x}} $

 What is the sum of the coefficients  of  $x$ after the first derivative and the second derivative of the function $2{{x}^{4}}-\sqrt{3}{{x}^{3}}+5x^2-x+6$

 To find the difference in between the coefficient of first and second derivative we will first find the derivatives and then subtract the coefficient obtained in both the cases.

  $f(x)=2{{x}^{4}}-\sqrt{3}{{x}^{3}}+5x^2-x+6 \\ \text{Differntiate wrt x} \\ f'(x)=8{{x}^{3}}-3\sqrt{3}{{x}^{2}}+10x-1 \\ \text{Coeffiecient of x when taking first derivative = 10} \\ \text{Differntiate wrt x} \\ f''(x)=24{{x}^{2}}-6\sqrt{3}x+10 \\ \text{Coeffiecient of x when taking second derivative = }6\sqrt{3} \\ \text{Sum of the coefficient of x: } 10+6\sqrt{3}$

 Find the second derivative of the function  $\dfrac{{{x}^{3}}}{\sqrt{3}}+{{e}^{2{{x}^{2}}}}-4$

 As mentioned in the question, to find the second derivative we need to differentiate the given function twice and then we get a second derivative.

  $f(x)=\dfrac{{{x}^{3}}}{\sqrt{3}}+{{e}^{2{{x}^{2}}}}-4 \\ \text{Differentiate wrt x} \\ f'(x)=\dfrac{3{{x}^{2}}}{\sqrt{3}}+{{e}^{2{{x}^{2}}}}\times 4x \\ \text{Again differentiate wrt x:} \\ f''(x)=2\sqrt{3}x+{{e}^{2{{x}^{2}}}}\times 4+{{e}^{2{{x}^{2}}}}\times 4x\times 4x \\ \Rightarrow 2\sqrt{3}x+4{{e}^{2{{x}^{2}}}}+16{{x}^{2}}{{e}^{2{{x}^{2}}}} \\ \Rightarrow 2\sqrt{3}x+4({{e}^{2{{x}^{2}}}}+4{{x}^{2}}{{e}^{2{{x}^{2}}}})$

 Find the ratio of the second derivatives of $\dfrac{\sin x}{4}-\sqrt{3}\cos 2x$ and  $\sin 2x$

 To find the ratio of the second derivatives we have to find the second derivatives of both the function and then find its ratio.

  $f(x)=\dfrac{\sin x}{4}-\sqrt{3}\cos 2x \\ \text{Differentiate wrt x:} \\ f'(x)=\dfrac{\cos x}{4}+2\sqrt{3}\sin 2x \\ \text{Differentiate wrt x:} \\ f''(x)=-\dfrac{\sin x}{4}+4\sqrt{3}\cos 2x \\ g(x)=\sin 2x \\ \text{Differentiate wrt x:} \\ g'(x)=2\cos 2x \\ \text{Differentiate wrt x:} \\ g''(x)=-4\sin 2x \\ \dfrac{f''(x)}{g''(x)}=\dfrac{-\dfrac{\sin x}{4}+4\sqrt{3}\cos 2x}{-4\sin 2x} \\ \Rightarrow \dfrac{\sin x}{32\sin x\cos x}-\sqrt{3}\cot 2x \\ \Rightarrow \dfrac{1}{32\cos x}-\sqrt{3}\cot 2x$

 Find the local maximum or local minimum of the function  $f(x)={{x}^{4}}+\dfrac{{{x}^{3}}}{3}-{{\left( x-12 \right)}^{2}}$

 Since we need to find the local maxima or minima therefore we check the value of $f'(x)$ if it has a maximum value with respect to the points nearby it then it is local maxima else minima.

  $f(x)=f(x)={{x}^{4}}+\dfrac{{{x}^{3}}}{3}-{{\left( x-12 \right)}^{2}} \\ \text{Differentiate this wrt x:} \\ f'(x)=4{{x}^{3}}+{{x}^{2}}-2\left( x-12 \right)=0 \\ \text{For finding maxima or minima }f'(x)=0 \\ \text{Therefore,} \\ x=-2 \\ \text{Again differentiate wrt x:} \\ f''(x)=12{{x}^{2}}+2x-2 \\ f''(-2)=12(4)-4-2 \\ \Rightarrow 42 \\ \text{Now since }f''(-2)\text{ is positive } \\ \text{Hence, point of local minima and the value at }x=-2\text{ is }42$