1) f '(x) = ((2x-3)⁵(3x²+2x+1))' = ((2x-3)⁵)'(3x²+2x+1) + (2x-3)⁵(3x²+2x+1)' = 10(2x-3)⁴(3x²+2x+1) + (2x-3)⁵(6x+2) = 2(2x-3)⁴(5(3x²+2x+1)+(2x-3)(3x+1))=2(2x-3)⁴(15x²+10x+5+6x²-7x-3)=2(2x-3)⁴(21x²+3x+2).
f '(1) = 52
2) f '(x) = (⁴√(3x+1)·(3x-14)⁴)' = (⁴√(3x+1))'·(3x-14)⁴+⁴√(3x+1)·((3x-14)⁴)' =
f '(5) =
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
1) f '(x) = ((2x-3)⁵(3x²+2x+1))' = ((2x-3)⁵)'(3x²+2x+1) + (2x-3)⁵(3x²+2x+1)' = 10(2x-3)⁴(3x²+2x+1) + (2x-3)⁵(6x+2) = 2(2x-3)⁴(5(3x²+2x+1)+(2x-3)(3x+1))=2(2x-3)⁴(15x²+10x+5+6x²-7x-3)=2(2x-3)⁴(21x²+3x+2).
f '(1) = 52
2) f '(x) = (⁴√(3x+1)·(3x-14)⁴)' = (⁴√(3x+1))'·(3x-14)⁴+⁴√(3x+1)·((3x-14)⁴)' =![\(\frac{1}{4\sqrt[4]{(3x+1)^3}}\cdot (3x-14)^4+\sqrt[4]{3x+1}\cdot 12(3x-14)^3=\\=\frac{(3x-14)^4+48(3x+1)(3x-14)^3}{4\sqrt[4]{(3x+1)^3}}=\frac{3(3x-14)^3((3x-14)+16(3x+1)}{4\sqrt[4]{(3x+1)^3}}=\\=\frac{3(3x-14)^3(3x-14+48x+16)}{4\sqrt[4]{(3x+1)^3}}=\frac{3(3x-14)^3(51x+2)}{4\sqrt[4]{(3x+1)^3}}\)](https://tex.z-dn.net/?f=%5C%28%5Cfrac%7B1%7D%7B4%5Csqrt%5B4%5D%7B%283x%2B1%29%5E3%7D%7D%5Ccdot%20%283x-14%29%5E4%2B%5Csqrt%5B4%5D%7B3x%2B1%7D%5Ccdot%2012%283x-14%29%5E3%3D%5C%5C%3D%5Cfrac%7B%283x-14%29%5E4%2B48%283x%2B1%29%283x-14%29%5E3%7D%7B4%5Csqrt%5B4%5D%7B%283x%2B1%29%5E3%7D%7D%3D%5Cfrac%7B3%283x-14%29%5E3%28%283x-14%29%2B16%283x%2B1%29%7D%7B4%5Csqrt%5B4%5D%7B%283x%2B1%29%5E3%7D%7D%3D%5C%5C%3D%5Cfrac%7B3%283x-14%29%5E3%283x-14%2B48x%2B16%29%7D%7B4%5Csqrt%5B4%5D%7B%283x%2B1%29%5E3%7D%7D%3D%5Cfrac%7B3%283x-14%29%5E3%2851x%2B2%29%7D%7B4%5Csqrt%5B4%5D%7B%283x%2B1%29%5E3%7D%7D%5C%29 "\(\frac{1}{4\sqrt[4]{(3x+1)^3}}\cdot (3x-14)^4+\sqrt[4]{3x+1}\cdot 12(3x-14)^3=\\=\frac{(3x-14)^4+48(3x+1)(3x-14)^3}{4\sqrt[4]{(3x+1)^3}}=\frac{3(3x-14)^3((3x-14)+16(3x+1)}{4\sqrt[4]{(3x+1)^3}}=\\=\frac{3(3x-14)^3(3x-14+48x+16)}{4\sqrt[4]{(3x+1)^3}}=\frac{3(3x-14)^3(51x+2)}{4\sqrt[4]{(3x+1)^3}}\)")
f '(5) =