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(i) Let f(x) = ` sqrt(3x -2)` <br> f(x) = ` 1/2 (3x -2)^(-1/2) d / (dx) (3x-2)` <br> ` = 1/2 (3x-2) ^(-1/2) (3)` <br> ` 3/(2sqrt(3x-2)) x lt 2/3` <br> (ii) ` y = (2x^(2) +3) ^(5/3) (x+5)^(-1/3)` <br> Differentiating w.r.t x , we get <br> `(dy)/(dx) = (2x^(2)+3)^(5/3) d/(dx) (x+5)^(-1/3) + (x+5)^(-1/3) d/(dx) ((2x^(2) +3)^(5/3))` <br> `= (2x^(2)+3)^(5/3)(-1/3)(x+5)^(-1/3-1) d/(dx) (x+5) + (x+5) + (x +5)^(-1/3) . 5/3(2x^(2)+3)^(5/3-1) d/(dx) (2x^(2)+3)` <br> `= (2x^(2)+3)^(5/3) (-1/3)(x+5)^(-4/3) 1+ (x=5)^(-1/3) .1 + (x+5)^(-1/3) 5/3 (2x^(2) +3)^(2/3)(4x)` <br> `1/3 (2x^(2)+3)^(2/3) (x+5)^(-1/3) (-2(2x^(2)+3)(x+5)^(-1) + 20x)` <br> ` 1/3(2x^(2)+3)^(2/3)(x+5)^(-1/3) ( 20x - (2x^(2)+3)/(x+5))` <br> `= 1/3 ((2x^(2)+3)^(2/3)(18x^(2)+100x-3))/((x+5)^(4/3)), (x ne -5)` <br> (x+5) Transcript

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00:00 - 00:59 | hello send the given problem is find the derivative of the following functions with respect to X so first function is given this and S is this to first win not first function so what is first function first given that f x is equals to let's say we are let this y y is equals to under the root 3 x minus 2 so what is the difference equation of root x type of function if root access their find the derivative of this so how to find the derivative of this worst we write this thing Express 2 part 1 by 2 and this is equals to 1 by 2 X raise to power 1 by 2 minus 1 equals to 1 by 2 into root x in the denominator so this is the answer now if you apply it over here is what is the DY by DX over here this device dx12 is under root root access 1 upon 2 root x this and DY by DX of disturb inside time is 3 x minus 2 so here we are using chain rule |

01:00 - 01:59 | when is become exposed to the definition of 3 x minus 2 root 3 root 3 by 2 X under root 3 x minus 2 this is the answer of the first problem now second is what the second part of the problem is here Y is equal to given is wise to 2 X square + 3 raise to the power 2 X square + 3 raise to the power 5 by 3 this is given 5 by 3 x 8 x + 3 raise to the power minus 1 by 3 this is the problem yes is it it is true yes it is through here is 5 so this is the problem S now this letter cost to this is equals to why so how to solve this type of problem first we are using their product rule now what is product rule the product rule se is the DY by DX of DY by DX of |

02:00 - 02:59 | function vichar product which are in the product this function we are using PX DY DX of q x + q x x x this is the founder of we are using over here so fast that this DY DX over here and this will be equal to first one said let's say this and this is qx PS and QS let's say this PX and this is our us to apply this first PS is it is should 2 X square + 3 x to the power 5 by 3 this is a Pappi DY DX of cubexo DY by DX of us value is this X + 5 raise to the power minus 1 by 3 + what is qxqzf4ywvys of qx PX what is p x 2 X square + 3 raise to the power 5 by 3 now simplify this so this value become |

03:00 - 03:59 | what this is a quest to first we write this this is as it is 2 x square + 3 raise to the power 5 by 3 + 1 then as it is in differentiation of this what this difference is minus 1 by 3 X + 5 raise to the power 1 by 3 minus 1 minus 4 by 3 + this is a product so here this is X + 5 raise to the power minus 1 by 3 and what is the DY DX of this this is 5 by 3 2 X square + 3 raise to the power 5 by 3 minus 12 this value is equal to 2 by 3 and DY by DX of this inside function so this is 4x 9 simplify and get the answer the simplified form of this this is equal to 2 X square + 3 raise to the power 5 by 3 minus 1 by 3 outside and hear this is X + 5 raise to the power minus 4 by 3 and this value is equal to what this here this is plus science this sport in 25 + 23 |

04:00 - 04:59 | X + 5 raise to the power minus 1 by 3 and 2 X square + 3 raise to the power 2 by 3 27000 final answer if you can simplify more then also we can more simplified but no need to more simplifying this so this is our final answer of the problem this is the final answer answer of the second part thank you |