Question

Find $\frac{dy}{dx}$, if y = log ( sec θ + tan θ), x = sec θ, at θ = π/4

Answer 1

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Answer 2

Answer:

$\dfrac 12$

Step-by-step explanation:

$y=log(sec\theta + tan\theta)$

On differentiating the equation we get,

$\dfrac {dy}{d\theta} = \dfrac {1}{sec\theta + tan \theta}$

Now put the value of $\theta = \pi/4 = 45^o$

$\dfrac {dy}{d\theta} = \dfrac {1}{sec 45^o + tan 45^o}\\\dfrac {dy}{d\theta} = \dfrac 1{1+1}\\\dfrac {dy}{d\theta} = \dfrac 12$ -----equation 1

$x = sec\theta$

On differentiating the equation we get,

$\dfrac {dx} {d\theta} = sec\theta. tan\theta$

Now put the value of $\theta = \pi/4 = 45^o$

$\dfrac {dx} {d\theta} = sec45^o. tan45^o\\\dfrac {dx} {d\theta}  = 1 \times 1 \\\dfrac {dx} {d\theta} = 1$  ------- equation 2

Now divide equation 1 by equation 2

$\dfrac {\dfrac {dy}{d\theta} }{\dfrac {dx}{d\theta}} = \dfrac{\dfrac 12}1$

$\dfrac {dy} {dx} = \dfrac 12$