Partial Differentiation || 𝒆^(𝒂𝒙+𝒃𝒚) 𝒇(𝒂𝒙−𝒃𝒚) || 22mat11 || 18mat21 || Dr Prashant Patil

  Рет қаралды 13,014

Dr Prashant Patil

Dr Prashant Patil

Күн бұрын

Пікірлер: 15
@AyushGaurav10
@AyushGaurav10 Ай бұрын
Thank you so much sir you explained it really well 🙏🏻
@praveentonytony612
@praveentonytony612 3 ай бұрын
Very nice explanation Steps explained clearly
@cartoonworld2729
@cartoonworld2729 Жыл бұрын
Sir your method of explaining very nice 🔥🔥🔥
@ManojKote-iq6up
@ManojKote-iq6up 3 ай бұрын
Super sir I understood 🎉❤
@princesssunshine710
@princesssunshine710 11 ай бұрын
Thank you so much for this explanation sir! ❤
@sujalmaradi5755
@sujalmaradi5755 9 ай бұрын
Thanks sir ❤
@guruprasad5189
@guruprasad5189 10 ай бұрын
Thanks sir
@Arpithays-m5q
@Arpithays-m5q 10 ай бұрын
🙏🏻
@raviuppar05
@raviuppar05 4 ай бұрын
🌟✔️
@okeybyby2283
@okeybyby2283 11 ай бұрын
❤🙏
@jithutech
@jithutech 9 ай бұрын
🙏🙏🙏🙏❤️❤️❤️
@Nami-hb4dw
@Nami-hb4dw 9 ай бұрын
First learn how to teach
@Syck834
@Syck834 9 ай бұрын
it is wrong in both the cases
@user-zk6xz
@user-zk6xz 9 ай бұрын
How?
@Syck834
@Syck834 9 ай бұрын
@@user-zk6xz after differentiating Ux= a.e^ax................... + ua will come and then we shld multiply with "b" on both the sides So it becomes b ux = abe^ax+by[f¹(ax - by)]+abu Same procedure with Uy
Partial Differentiation || 𝒇(𝒙+𝒄𝒕)+𝒈(𝒙−𝒄𝒕) || 22mat11 || 18mat21 || Dr Prashant Patil
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