Question

find dy/dx when y= 1/(secx-tanx)

Answer 1

$\huge{ \red{ \mathfrak{solution}}}$

Given:

$\boxed{ \bold{y \: = \frac{1}{secx \: - tanx} }}$

To find :

dy/dx

• First solve y

$\star \: y = \frac{1}{secx - tanx} \\ \\ \rightarrow \: y = \frac{1}{secx - tanx} \times \frac{secx + tanx}{secx + tanx} \\ \\ \rightarrow y = \frac{secx + tanx}{ {sec}^{2}x - {tan}^{2} x }$

As we know that

•sec^2@-tan^2@ = 1

$\rightarrow y = \: secx + tanx$

Now differentiate w.r.t X

$\implies \frac{d}{dx} (secx + tanx) \\ \\ \implies \frac{d}{dx} (secx) + \frac{d}{dx} (tanx) \\ \\ \implies \: secx \: tanx \: + {sec}^{2} x \\ \\ \implies \: secx(tanx + secx)$

Formula used :

$\red{ \star} \boxed{ \bold{ \frac{d}{dx} secx = secx \: tanx}} \\ \\ \red\star \boxed{ \bold{\frac{d}{dx} tanx = {sec}^{2} x}}$

Answer 2

Answer:

$formula \: used \: =$

$|| \frac{d}{dx}secx \: = \: secx \: tanx \: .$

$|| \frac{d}{dx}tanx \: = sec {}^{2} x.$

$to \: find \: = dy \: \:and \: dx.$

$first \: we \: can \: solve \: y \: .$

$y \ = \frac{1}{secx \: - \: tanx \: }$

$y \times \frac{1}{secx \: - \: tanx \: } \times \frac{secx \: + \: tanx \: }{secx \: + \: tanx \: }$

$y \times \frac{secx \: + \: tanx}{ {sec}^{2} x \: - tan^{2} x}$

$therefore \: we \: know \: that \: =$

${sec}^{2}at \: the \: rate \: - \: tan {}^{2}at \: the \: rate \: = \: 1.$

$y \: = \: secx + tanx \: .$

$now \: differntiate \: w.r.t \: x.$

$\frac{d}{dx} (secx + tanx).$

$\frac{d}{dx} (secx \: + \frac{d}{dx} \: ( \: tanx \: ) \: .$

$secx \: tanx \: + \: sec {}^{2}x.$

$secx \: ( \: tanx \: + \: secx \: ) .$