Math, asked by sakshiii39, 11 months ago

find the derivative tanx+cotx/logx​

Answers

Answered by gauravprakash231
2

Step-by-step explanation:

use u by v rule means u ko diff karo v ko vaise hi rhne do then minus karo aur phir v ko diff karo u ko vaise hi rhne do fir dono ka by v^2 karo

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Answered by tiwariakdi
0

The derivative of the function tanx + cotx/logx is sec^2x - (csc^2x/ logx) - cotx/(logx * x^2).

To find the derivative of the given function, we need to use the sum and quotient rules of differentiation.

Let f(x) = tanx + cotx/logx

Using the sum rule, we have:

f'(x) = (tanx)' + (cotx/logx)'

Now, to differentiate tanx , we use the chain rule:

(tanx)' = sec^2x

And to differentiatecotx/logx, we use the quotient rule

(cotx/logx)' = [(cotx)'(logx) - (cotx)(logx)'] / (logx)^2

= [-csc^2x(logx) - cotx(1/x)] / (logx)^2

= -csc^2x(logx)/ (logx)^2 - cotx/[(logx)^2 * x]

Now, substituting these into f'(x), we get:

f'(x) = sec^2x - csc^2x(logx)/ (logx)^2 - cotx/[(logx)^2 * x]

Simplifying this expression, we get:

f'(x) = sec^2x - (csc^2x/ logx) - cotx/(logx * x^2)

Therefore, the derivative of the function tanx + cotx/logx is sec^2x - (csc^2x/ logx) - cotx/(logx * x^2).

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