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MTH101 GDB Solution & Discussion Due Date: 13 Dec. 2016
GDB No.1 (Lecture No. 13 to 19)
Question:
Please click here for Question of GDB No. 1.
Starting Date: Thursday, December 8, 2016, at 12:00 A.M.
Closing Date: Tuesday, December 13, 2016 at 11:59 P.M.
Ø Post your solution of GDB only on GDB interface of LMS.
Ø Do not post the solution of GDB on ‘regular MDB’ forum. It will not be graded on MDB.
Ø There is preview option available on GDB forum. So, see the preview of your solution before posting your GDB.
Ø Once GDB is posted, it cannot be editable or re-submitted.
Ø In case, you see boxes instead of your solution in your GDB interface then email your GDB solution at mth101@vu.edu.pk till due date.
The following will be helpful to you for inserting mathematical equations in GDB interface:
BEST OF LUCK!
Tags:
This Content Originally Published by a member of VU Students.
Views: 2749
kia ye ans bilkul sahi hai?
billkul lekin tareqa lengthy hay. iska answer sirf 3 lines me agr ap ne product rule, trignometric derivativs parhy hoay hain
for convenience view.... Anyone need help in MTH101... I am available every time Guys.... :)
GDB main post kesy karna h plzzzzzz guide kr den
1. Install mathtype in your computer and Open it:
2. Go to the following path in Menu bar:
Preferences > Cut and Copy Preferences
3. Dialog Box me MathML or TeX ko check kren or is k neechay dropdown list me se AMSLaTeX ko select kren. Or is k ilawa baqi sb ko Uncheck kr den.
4. Ab apna answer type kren
5. Mouse se select kr k Right Click kr Copy kr len.
6. GDB post comment me ja k Apna message post kr k apni equation paste kren.
7. Neechay Preview me preview check kren agr preview sahe to post kr den Otherwise apni answer equation ko dobara check kren, is me kahen koi space to nhi hay? or dobara copy kr k paste kr k preview dekh k post kr den.
math type 6.6 ya 6.9 ya simple mathtype
6.9 is updated version and is also better
\[\begin{gathered}
y = 5\tan \sqrt {x\,} \hfill \\
Solution: - \hfill \\
Chain\,\,\,rule\,\,...........\,\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}.\frac{{du}}{{dx}}..............\left( A \right) \hfill \\
\,Let\,\,u = \sqrt {x\,} \hfill \\
y = 5\tan u\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & & | & u = \sqrt {x\,} \hfill \\
taking\,\,derivative\,\,w.r.t\,\,\,\,\,(u) &&& \frac{{du}}{{dx}} = \frac{1}{{2\sqrt x }}.\frac{d}{{dx}}(x) \hfill \\
\frac{{dy}}{{du}} = 5{\sec ^2}u.\frac{d}{{du}}(u) & & & \frac{{du}}{{dx}} = \frac{1}{{2\sqrt x }}.1 \hfill \\
\frac{{dy}}{{du}} = 5{\sec ^2}u.1 & & & \frac{{du}}{{dx}} = \frac{1}{{2\sqrt x }}...........(2) & & & & & & \hfill \\
\frac{{dy}}{{du}} = 5{\sec ^2}u............(1) \hfill \\
putting\,\,the\,\,value\,\,of\frac{{dy}}{{du}} = 5{\sec ^2}u\,\,\& \,\,\frac{{du}}{{dx}} = \frac{1}{{2\sqrt x }}\,\,in\,\,eq(A) \hfill \\
& & & \hfill \\
& \frac{{dy}}{{dx}} = \frac{{dy}}{{du}}.\frac{{du}}{{dx}} \hfill \\
\,\,\,\, & \frac{{dy}}{{dx}} = 5{\sec ^2}u\,.\frac{1}{{2\sqrt x }} \hfill \\
& \frac{{dy}}{{dx}} = \frac{{5{{\sec }^2}u\,}}{{2\sqrt x }}\,\,\,\,\,\,\,\therefore u = \sqrt x \hfill \\
& \frac{{dy}}{{dx}} = \frac{{5{{\sec }^2}\sqrt x }}{{2\sqrt x }}\,\, \hfill \\
\hfill \\
\end{gathered} \]% MathType!End!2!1!
Dear Students,
Please, download the attached file for the guidelines that how to upload your GDB of any MTH Subject.
Thank You
&
Best of Luck!
required in GDB
1...common ratio
2...using formula
3...and answer of sum is
series
7 numbers of series will be
a1,a2,a3,a4,a5,a6,a7
1-2+4-8+16-32+64
Sum of 7 numbers = 43
r means the value that is multiplied with old number to make a new number
like 1st number = 1 * -2 =-2 (2nd number)
2nd numebr = -2 * -2 =4 (3rd number
3rd number = 4 * -2 = -8 (4th number )
i mean to say that 2 is the value that is producing new series number so r= -2
a7= a1(r)^n-1
a7= 1 (-2)^7-1
a7 = 2^6
a7 = 64
to find 6th value using formula , apply this formula again
common ratio sab mai same hotii hai as proved below
common ratio = -2 (if i m nt wrng)
example =
common ratio = a2/a1 = -2/1 = -2
= a3/a2 = 4/-2 = -2
= a4/a3 = -8/4 = -2
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