{"id":3331,"date":"2023-12-07T04:14:08","date_gmt":"2023-12-06T22:44:08","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=3331"},"modified":"2023-12-08T17:54:50","modified_gmt":"2023-12-08T12:24:50","slug":"samacheer-kalvi-11th-maths-solutions-chapter-10-ex-10-3","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-11th-maths-solutions-chapter-10-ex-10-3\/","title":{"rendered":"Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.3"},"content":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions<\/a> Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.3<\/h2>\n

Differentiate the following
\nQuestion 1.
\ny = (x2<\/sup> + 4x + 6)5<\/sup>
\nSolution:
\nLet = u = x2<\/sup> + 4x + 6
\n\u21d2 \\(\\frac{d u}{d x}\\) = 2x + 4
\nNow y = u5<\/sup> \u21d2 \\(\\frac{d y}{d x}\\) = 5u4<\/sup>
\n\u2234 \\(\\frac{d y}{d x}=\\frac{d y}{d u} \\times \\frac{d u}{d x}\\) = 5u4<\/sup> (2x + 4)
\n= 5(x2<\/sup> + 4x + 6)4<\/sup> (2x + 4)
\n= 5 (2x + 4) (x2<\/sup> + 4x + 6)4<\/sup><\/p>\n

Question 2.
\ny = tan 3x
\nSolution:
\ny = tan 3x
\nput u = 3x
\n\\(\\frac{d u}{d x}\\) = 3
\nNow y = tan u
\n\u21d2 \\(\\frac{d u}{d x}\\) = sec2<\/sup> u
\nSo \\(\\frac{d y}{d x}=\\frac{d y}{d u} \\times \\frac{d u}{d x}\\) = (sec2<\/sup> u) (3)
\n= 3 sec2<\/sup> 3x<\/p>\n

Question 3.
\ny = cos (tan x)
\nSolution:
\nPut u = tan x
\n\\(\\frac{d u}{d x}\\) = sec2<\/sup>x
\nNow y = cos u \u21d2 \\(\\frac{d u}{d x}\\) = – sin u
\nNow \\(\\frac{d y}{d x}=\\frac{d y}{d u} \\times \\frac{d u}{d x}\\)
\n= (-sin u) (sec2<\/sup>x)
\n= – sec2<\/sup> (sin (tan x))<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\ny = \\(\\sqrt[3]{1+x^{3}}\\)
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 5.
\ny = \\(e^{\\sqrt{x}}\\)
\nSolution:
\n\"Samacheer<\/p>\n

Question 6.
\ny = sin (ex<\/sup>)
\nSolution:
\ny = sin (ex<\/sup>)
\nLet u = ex<\/sup>
\n\"Samacheer<\/p>\n

Question 7.
\nF(x) = (x3<\/sup> + 4x)7<\/sup>
\nSolution:
\nF(x) = (x3<\/sup> + 4x)7<\/sup>
\nPut u = x3<\/sup> + 4x
\n\"Samacheer<\/p>\n

Question 8.
\nh(t) = \\(\\left(t-\\frac{1}{t}\\right)^{\\frac{3}{2}}\\)
\nSolution:
\n\"Samacheer<\/p>\n

Question 9.
\nf(t) = \\(\\sqrt[3]{1+\\tan t}\\)
\nSolution:
\n\"Samacheer<\/p>\n

Question 10.
\ny = cos (a3<\/sup> + x3<\/sup>)
\nSolution:
\n\"Samacheer<\/p>\n

Question 11.
\ny = e-mx<\/sup>
\nSolution:
\n\"Samacheer<\/p>\n

Question 12.
\ny = 4 sec 5x
\nSolution:
\n\"Samacheer<\/p>\n

Question 13.
\ny = (2x – 5)4<\/sup> (8x2<\/sup> – 5)-3<\/sup>
\nSolution:
\n\"Samacheer
\n= \\(\\frac{8(2 x-5)^{3}}{\\left(8 x^{2}-5\\right)^{4}}\\) (-4x2<\/sup> + 30x – 5)<\/p>\n

\"Samacheer<\/p>\n

Question 14.
\ny = (x2<\/sup> + 1) \\(\\sqrt[3]{x^{2}+2}\\)
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 15.
\ny = xe-x2<\/sup><\/sup>
\nSolution:
\ny = xe-x2<\/sup><\/sup>
\ny = uv where u = x and v = e-x2<\/sup><\/sup>
\nNow u’ = 1 and v\u2019 = e-x2<\/sup><\/sup> (-2x)
\nv’ = – 2xe-x2<\/sup><\/sup>
\nNow y = uv \u21d2 y\u2019 = uv\u2019 + vu\u2019
\n(i.e.) \\(\\frac{d y}{d x}\\) = x[-2xe-x2<\/sup><\/sup>] + e-x2<\/sup><\/sup> (1)
\n= e-x2<\/sup><\/sup> (1 – 2x2<\/sup>)<\/p>\n

Question 16.
\ns(t) = \\(\\sqrt[4]{\\frac{t^{3}+1}{t^{3}-1}}\\)
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 17.
\nf(x) = \\(\\frac{x}{\\sqrt{7-3 x}}\\)
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 18.
\ny = tan (cos x)
\nSolution:
\ny = tan (cos x)
\n\"Samacheer<\/p>\n

Question 19.
\ny = \\(\\frac{\\sin ^{2} x}{\\cos x}\\)
\nSolution:
\n\"Samacheer<\/p>\n

Question 20.
\ny = \\(5^{-\\frac{1}{x}}\\)
\nSolution:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 21.
\ny = \\(\\sqrt{1+2 \\tan x}\\)
\nSolution:
\n\"Samacheer<\/p>\n

Question 22.
\ny = sin3<\/sup>x + cos3<\/sup>x
\nSolution:
\ny = sin3<\/sup>x + cos3<\/sup>x
\nHere u = sin3<\/sup> x = (sin x)3<\/sup>
\n\u21d2 \\(\\frac{d u}{d x}\\) = 3 (sin x)2<\/sup> (cos x)
\n= 3sin2<\/sup>x cos x
\nv = cos3<\/sup>x = (cos x)3<\/sup>
\n\u21d2 \\(\\frac{d v}{d x}\\) = 3 (cos x)2<\/sup> (-sin x) = -3 sin x cos2<\/sup>x
\nNow y = u + v \u21d2 \\(\\frac{d y}{d x}=\\frac{d u}{d x}+\\frac{d v}{d x}\\)
\n= 3 sin2<\/sup>x cos x – 3sin x cos2<\/sup>x
\n= 3 sin x cos x (sin x – cos x)<\/p>\n

Question 23.
\ny = sin2<\/sup> (cos kx)
\nSolution:
\n\"Samacheer<\/p>\n

Question 24.
\ny = (1 + cos2<\/sup>x)6<\/sup>
\nSolution:
\n\"Samacheer<\/p>\n

Question 25.
\ny =\\(\\frac{e^{3 x}}{1+e^{x}}\\)
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 26.
\ny = \\(\\sqrt{x+\\sqrt{x}}\\)
\nSolution:
\n\"Samacheer<\/p>\n

Question 27.
\ny = ex cos x<\/sup>
\nSolution:
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 28.
\ny = \\(\\sqrt{x+\\sqrt{x+\\sqrt{x}}}\\)
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n

Question 29.
\ny = \\(\\sin (\\tan (\\sqrt{\\sin x}))\\)
\nSolution:
\n\"Samacheer<\/p>\n

Question 30.
\ny = sin-1<\/sup>\\(\\left(\\frac{1-x^{2}}{1+x^{2}}\\right)\\)
\nSolution:
\n\"Samacheer
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.3 Differentiate the following Question 1. y = (x2 + 4x + 6)5 […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[7],"tags":[],"blocksy_meta":"","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/3331"}],"collection":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/comments?post=3331"}],"version-history":[{"count":1,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/3331\/revisions"}],"predecessor-version":[{"id":55203,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/3331\/revisions\/55203"}],"wp:attachment":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/media?parent=3331"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/categories?post=3331"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/tags?post=3331"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}