Math Assignment Class XII Ch-5 | Derivatives

Math Assignment Class XII Chapter 5 Derivatives

 Question 1 Use chain rule to find the derivative of \[y=\left ( \frac{2x-1}{2x+1} \right )^{2} \;\;\; Ans.1: \frac{8(2x-1)}{(2x+1)^{3}}\]

Question 2. Differentiate the following w.r.t. x \[y=log_{10}x+ log_{x}10+log_{x}x+ log_{10}10 \] Answer 2 \[ \frac{dy}{dx}=\frac{1}{xlog10}-\frac{log10}{x(logx)^{2}} \] Hint for the solution: \[\frac{logx}{log10}+\frac{log10}{logx}+\frac{logx}{logx}+\frac{log10}{log10} \]\[=\frac{logx}{log10}+\frac{log10}{logx}+1+1\] Now differentiating w. r. t. x and taking log10 as constant.

Question 3. Differentiate the following w. r. t. x at x = 1\[y=e^{x(1+logx)}\; \; \; \; \; \; \; \;........\: \: Ans.[2e]\]

Question 4. \[ If\: \frac{x}{x-y}=log\frac{a}{x-y},\: then\: prove\: that\: \frac{dy}{dx}=\frac{2y-x}{y}\]

Question 5: Differentiate log(x e^x)  w. r. t.   xlogx. 

Answer 5: \[ \frac{1+x}{x(1+logx)}\]

Question 6: Differentiate x² w. r. t. x³.        ...........    Ans:  2/3x.

Question 7: \[ If\: e^{x}+e^{y}=e^{x+y}, prove\: that\: \: \frac{dy}{dx}+e^{y-x}=0\] Solution 7: \[ e^{x}+e^{y}=e^{x+y}\\\frac{e^{x}}{e^{x+y}}+\frac{e^{y}}{e^{x}+e^{y}}=1\Rightarrow e^{-y}+e^{-x}=1\]Now differentiating w. r. t. x we get the required answer.

Question 8: Differentiate  (x^x)^x  w. r. t. x  

Answer 8: \[ \frac{dy}{dx}=x^{x^{2}}[x+2xlogx]...[Hint:(x^{x})^{x}=x^{x^{2}}]\] 

Question9.   Differentiate x¹⁶y⁹ = (x² + y)¹⁷ w. r. t. x ,

Answer 9:  \[ \frac{dy}{dx}=\frac{2y}{x}\]

Question10) Differentiate  (logx)^x + x^logx, w. r. t. x. 

Answer 10: \[ (logx)^{x}\left [ \frac{1}{logx}+log(logx) \right ]+2x^{logx-1}.logx\]

Question11\[If\: \: f(x)=|cosx|, find\: f'\left ( \frac{3\pi }{4} \right )...Ans.\frac{1}{\sqrt{2}}\]

Question 12: \[If\: \: y=\sqrt{cosx+\sqrt{cosx+\sqrt{cos+....\infty }}},\: \\prove\: that\: (2y-1)\frac{dy}{dx}=-sinx\]Solution 12:\[Since\: \: y=\sqrt{cosx+\sqrt{cosx+\sqrt{cos+....\infty }}},\]\[y=\sqrt{cosx+y}\: \: \Rightarrow \: \: y^{2}=cosx+y\]Differentiating both side w. r. t. x we get \[2y\frac{dy}{dx}=-sinx+\frac{dy}{dx}\]\[(2y-1)\frac{dy}{dx}=-sinx\]

Question 13. Find dy/dx at x = 1, y = π/4, If Sin2y + cosxy = k
Answer 13.\[ \frac{\pi (\sqrt{2}+1)}{4}\]
Question 14. If x = a(2θ - sin2θ) and y = a(1-cos2θ), find dy/dx, when θ = π/4.
Answer 14: [ 1]
Question 15. Differentiate Log(cosex) w. r. t x ......... Ans[-ex tanx].
Question 16.\[If\: y=\sqrt{tanx+\sqrt{tanx+\sqrt{tanx+......\infty }}} \]\[Prove\: that, (2y-1)\frac{dy}{dx}=Sec^{2}x\] 
Question 17. \[If\: y=\frac{sin^{-1}x}{\sqrt{1-x^{2}}},show\: that:\] \[(1-x^{2})\frac{d^{2}y}{dx^{2}}-3x\frac{dy}{dx}-y=0\]
Question 18. \[If\: x=cost+log\: tan\frac{t}{2},\: y=sint, then \]\[ find\: the\: value\: of\: \: \frac{d^{2}y}{dx^{2}}\: at\: \: t=\frac{\pi }{4}\]
Question 19: Differentiate the following w. r. t. x\[y=sin^{-1}\left ( \frac{2^{x+1}}{1+4^{x}} \right )\: .....\: \: Ans:\: \: \frac{2^{x+1}}{1+4^{x}}log\: 2\]
Question 20. Differentiate the following w. r. t. x \[y=sin^{-1}\left \{ \frac{2^{x+1}.3^{x}}{1+(36)^{x}} \right \}.\: \: ....\: Ans:\: \frac{2(log6)6^{x}}{1+(36)^{x}}\]
Question 21. \[Differentiate\: y=e^{e^{x}}\: w.\: r.\: t.\: x.\: \: Ans:\: e^{e^{x}}\times e^{x}\]
Question 22. Find the value of k if the following function is continuous.\[f(x)=\left\{\begin{matrix} \frac{kcosx}{\pi -2x}\; \; ,\; \; x<\frac{\pi }{2}\\\\\frac{3tan2x}{2x-\pi }\; \; ,\; \; x>\frac{\pi }{2} \end{matrix}\right.\; \; \; \; \; \; \; \: \: \: ....\: \: Ans\: \: k=6\]
Question 23. Find the value of k for which the following function is continuous at x = 0\[f(x)=\left\{\begin{matrix} \frac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\; \; ,\; \; -1\leq x<0\\\\ \frac{2x+1}{x-2}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \:, \: \: \: \: \: \: 0\leq x<1 \end{matrix}\right.\: \: \: \: \: .....\: \: \: \: \: \: Ans\: k=-\frac{1}{2}\]
Question 24: Show that f(x) = |x - 2| is continuous but not differentiable at x = 3.
Question 25: Discuss the continuity and differentiability of the following function. \[f(x)=\left\{\begin{matrix} 1-x\: \:\: \: \: \: \: \: ,\: \: x<1\\(1-x)(2-x),\: 1\leq x\leq 2 \\ 3-x\: \: \: \: \: ,\: \: \: \: \: x>2 \end{matrix}\right.\]
Answer 25: f(x) is continuous at x = 1 but not continuous at x = 2
f(x) is differentiable at x = 1.
As f(x) is not continuous at x = 2 so f(x) is not differentiable at x = 2.

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