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PSGCAS is emerged as the top ranked Arts colleges in Coimbatore followed by other colleges. Candidates can see the table below for top Arts colleges in Coimbatore with NIRF ranking -

Source: Official site and may vary.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhata{x}^2+2hxy+b{y}^2+2gx+2fy+c=0\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}\left(a{x}^2+2hxy+b{y}^2+2gx+2fy+c\right)=\frac{d}{dx}\left(0\right)\\ \Rightarrow a.2x+2h\left(x.\frac{dy}{dx}+y.1\right)+b.2y.\frac{dy}{dx}+2g.1+2f.\frac{dy}{dx}+0=0\\ \Rightarrow 2ax+2hx.\frac{dy}{dx}+2hy+2by.\frac{dy}{dx}+2g+2f.\frac{dy}{dx}=0\\ \Rightarrow 2hx.\frac{dy}{dx}+2by.\frac{dy}{dx}+2f.\frac{dy}{dx}=-2ax-2hy-2g\\ \Rightarrow \left(2hx+2by+2f\right).\frac{dy}{dx}=-2\left(ax+hy+g\right)\\ \Rightarrow \frac{dy}{dx}=\frac{-2\left(ax+hy+g\right)}{2\left(hx+by+f\right)}=\frac{-\left(ax+hy+g\right)}{\left(hx+by+f\right)}\\ \Rightarrow \frac{dy}{dx}=\frac{-\left(ax+hy+g\right)}{\left(hx+by+f\right)}\\ Now,differentiatingbothsidesw.r.t.y\\ \frac{d}{dy}\left(a{x}^2+2hxy+b{y}^2+2gx+2fy+c\right)=\frac{d}{dy}\left(0\right)\\ \Rightarrow 2ax.\frac{dx}{dy}+2h\left(y.\frac{dx}{dy}+x.1\right)+2by+2g.\frac{dx}{dy}+2f.1+0=0\\ \Rightarrow 2ax.\frac{dx}{dy}+2hy.\frac{dx}{dy}+2hx+2by+2g.\frac{dx}{dy}+2f=0\\ \Rightarrow 2ax\frac{dx}{dy}+2hy.\frac{dx}{dy}+2g.\frac{dx}{dy}=-2hx-2by-2f\\ \Rightarrow \left(2ax+2hy+2g\right).\frac{dx}{dy}=-2\left(hx+by+f\right)\\ \Rightarrow \frac{dx}{dy}=\frac{-2\left(hx+by+f\right)}{2\left(ax+hy+g\right)}=\frac{-\left(hx+by+f\right)}{\left(ax+hy+g\right)}\\ \Rightarrow \frac{dx}{dy}=\frac{-\left(hx+by+f\right)}{\left(ax+hy+g\right)}\\ \therefore \frac{dy}{dx}.\frac{dx}{dy}=\left[\frac{-\left(ax+hy+g\right)}{\left(hx+by+f\right)}\right].\left[\frac{-\left(hx+by+f\right)}{\left(ax+hy+g\right)}\right]=1\\ Hence,\frac{dy}{dx}.\frac{dx}{dy}=1.Hence,proved.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat{\left(x^2+y^2\right)}^2=xy\\ \Rightarrow x^4+y^4+2x^2{y}^2=xy\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}\left(x^4\right)+\frac{d}{dx}\left(y^4\right)+2.\frac{d}{dx}\left(x^2{y}^2\right)=\frac{d}{dx}\left(xy\right)\\ \Rightarrow 4x^3+4y^3.\frac{dy}{dx}+2\left[x^2.2y.\frac{dy}{dx}+y^2.2x\right]=x.\frac{dy}{dx}+y.1\\ \Rightarrow 4x^3+4y^3.\frac{dy}{dx}+4x^{2}y.\frac{dy}{dx}+4x{y}^2=x.\frac{dy}{dx}+y\\ \Rightarrow 4y^3.\frac{dy}{dx}+4x^{2}y.\frac{dy}{dx}-x.\frac{dy}{dx}=y-4x^3-4x{y}^2\\ \Rightarrow \left[4y^3+4x^{2}y-x\right]\frac{dy}{dx}=y-4x^3-4x{y}^2\\ \Rightarrow \frac{dy}{dx}=\frac{y-4x^3-4x{y}^2}{4y^3+4x^{2}y-x}\\ Hence,\frac{dy}{dx}=\frac{y-4x^3-4x{y}^2}{4y^3+4x^{2}y-x}\end{array}$

Culinary Arts courses entail the culinary arts of food preparation, cooking, and food presentation, usually in the form of meals. It includes various aspects such as Bakery & Confectionery, Patisserie, Wine Culinary, Culinary Management. Culinary Arts courses are offered at various levels such as UG, PG, Certification, Diploma, and PG Diploma. tudents can check the list of career options available after completing BA in Culinary Arts from Bunts Sangha's Ramanath Payyade College of Hospitality Management Studies:

Candidates can confirm their seats after selection by paying the Sant Baba Bhag Singh University fees. The total tuition fees for the BA programme is INR 89,650. However, the fee structure may include other components, such as security deposit, application fee, admission fee, etc. The mentioned fee amount is as per the official website/ sanctioning body. It is still subject to changes and hence, is indicative.

To secure a BSc seat at Mohamed Sathak College, candidates must first meet the basic eligibility criteria by passing Class 12 with the required Science subjects. Since admission is based on academic merit, students with higher marks have a better chance of selection. Submitting the application on time and ensuring all documents are complete also improves admission prospects.

According to NIRF ranking 2025, under "Colleges category", PSG College of Arts and Science - PSGCAS emerged to be as the top ranked Arts college in Coimbatore, secured 11th position followed by Sri Krishna Arts and Science and Sri Ramakrishna College of Arts and Science.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatta{n}^{-1}\left(x^2+y^2\right)=a\\ \Rightarrow x^2+y^2=tana\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}\left(x^2+y^2\right)=\frac{d}{dx}\left(tana\right)\\ \Rightarrow 2x+2y.\frac{dy}{dx}=0\Rightarrow 2y.\frac{dy}{dx}=-2x\\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{2y}=\frac{-x}{y}\\ Hence,\frac{dy}{dx}=\frac{-x}{y}\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatsec\left(x+y\right)=xy\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}sec\left(x+y\right)=\frac{d}{dx}\left(xy\right)\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right).\frac{d}{dx}\left(x+y\right)=x.\frac{dy}{dx}+y.1\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right).\left(1+\frac{dy}{dx}\right)=x.\frac{dy}{dx}+y\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right)+sec\left(x+y\right).tan\left(x+y\right).\frac{dy}{dx}=x.\frac{dy}{dx}+y\\ \Rightarrow sec\left(x+y\right).tan\left(x+y\right).\frac{dy}{dx}-x.\frac{dy}{dx}=y-sec\left(x+y\right).tan\left(x+y\right)\\ \Rightarrow \left[sec\left(x+y\right).tan\left(x+y\right)-x\right]\frac{dy}{dx}=y-sec\left(x+y\right).tan\left(x+y\right)\\ \Rightarrow \frac{dy}{dx}=\frac{y-sec\left(x+y\right).tan\left(x+y\right)}{sec\left(x+y\right).tan\left(x+y\right)-x}\\ Hence,\frac{dy}{dx}=\frac{y-sec\left(x+y\right).tan\left(x+y\right)}{sec\left(x+y\right).tan\left(x+y\right)-x}\end{array}$

Yes, Ambalika Institute of Management and Technology provides summer internship opportunities to its students. Check out the table below to know more about the number of students placed during Ambalika Institute of Management and Technology summer internships:

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhatsinxy+\frac{x}{y}=x^2-y\\ Differentiatingbothsidesw.r.t.x\\ \frac{d}{dx}sin\left(xy\right)+\frac{d}{dx}\left(\frac{x}{y}\right)=\frac{d}{dx}\left(x^2\right)-\frac{d}{dx}\left(y\right)\\ \Rightarrow cosxy.\frac{d}{dx}\left(xy\right)+\frac{y.\frac{d}{dx}\left(x\right)-x.\frac{dy}{dx}}{y^2}=2x-\frac{dy}{dx}\\ \Rightarrow cosxy\left[x.\frac{dy}{dx}+y.1\right]+\frac{y.1}{y^2}-\frac{x}{y^2}\frac{dy}{dx}=2x-\frac{dy}{dx}\\ \Rightarrow xcosxy.\frac{dy}{dx}+ycosxy+\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}=2x-\frac{dy}{dx}\\ \Rightarrow xcosxy.\frac{dy}{dx}-\frac{x}{y^2}\frac{dy}{dx}+\frac{dy}{dx}=2x-ycosxy-\frac{1}{y}\\ \Rightarrow \left[xcosxy-\frac{x}{y^2}+1\right]\frac{dy}{dx}=2x-ycosxy-\frac{1}{y}\\ \Rightarrow \left[\frac{x{y}^{2}cosxy-x+y^2}{y^2}\right]\frac{dy}{dx}=\frac{2xy-y^{2}cosxy-1}{y}\\ \Rightarrow \frac{dy}{dx}=\frac{2xy-y^{2}cosxy-1}{y}\times \frac{y^2}{x{y}^{2}cosxy-x+y^2}\\ =\frac{2x{y}^2-y^{3}cos\left(xy\right)-y}{x{y}^{2}cos\left(xy\right)-x+y^2}\\ Hence,\frac{dy}{dx}=\frac{2x{y}^2-y^{3}cos\left(xy\right)-y}{x{y}^{2}cos\left(xy\right)-x+y^2}\end{array}$