Question

find the derivative tanx+cotx/logx​

Answer 1

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

Answer 2

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

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 differentiate$cotx/logx,$ 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 t$anx + cotx/logx is sec^2x - (csc^2x/ logx) - cotx/(logx * x^2).$

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