Question

Answer the above question
"28th question".
#Class 12th
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Answer 1

EXPLANATION.

If cos(y) = x cos(a + y) with cos(a) ≠ ± 1.

As we know that,

We can write equation as,

⇒ cos(y)/cos(a + y) = x.

Differentiate w.r.t x, we get.

$\sf \implies x = \dfrac{cos (y)}{cos(a + y)}$

$\sf \implies \dfrac{d(x)}{dx} = \dfrac{d\bigg(\dfrac{cos(y)}{cos(a + y)}\bigg) }{dx}$

$\sf \implies 1 = \dfrac{d\bigg(\dfrac{cos(y)}{cos(a + y)} \bigg)}{dx} . \dfrac{dy}{dy}$

$\sf \implies 1 = \dfrac{d \bigg(\dfrac{cos(y)}{cos(a + y)} \bigg)}{dy} .\dfrac{dy}{dx}$

As we know that,

Division rule Or quotient rule.

$\sf \implies \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg] \ = \dfrac{g(x). \dfrac{d[f(x)]}{dx} \ - f(x). \dfrac{d[g(x)]}{dx} }{[g(x)]^{2} }$

Using this formula in equation, we get.

$\sf \implies 1 = \dfrac{cos(a + y). \dfrac{d[cos(y)]}{dy} \ - cos(y). \dfrac{d[cos(a + y)]}{dy} }{[cos(a + y)]^{2} } \ \times \dfrac{dy}{dx}$

$\sf \implies 1 = \dfrac{-sin(y) . cos(a + y) \ - cos(y). (- sin(a + y)) . \dfrac{d[a + y]}{dy} }{cos^{2} (a + y)} \ \times \dfrac{dy}{dx}$

$\sf \implies 1 = \dfrac{- sin(y) . cos(a + y) + cos(y)(sin(a + y)}{cos^{2}(a + y)} \ \times \dfrac{dy}{dx}$

$\sf \implies 1 = \dfrac{cos(y).sin(a + y) \ - sin(y).cos(a + y)}{cos^{2}(a + y) } \ \times \dfrac{dy}{dx}$

$\sf \implies 1 = \dfrac{sin\bigg(a + y) - y\bigg)}{cos^{2}(a + y )} \ \times \dfrac{dy}{dx}$

$\sf \implies 1 = \dfrac{sin(a)}{cos^{2}(a + y) } \ \times \dfrac{dy}{dx}$

$\sf \implies \dfrac{dy}{dx} \ = \dfrac{cos^{2} (a + y)}{sin(a)}$

                                                                                                                       

MORE INFORMATION.

Differentiation of f(x), g(x) type function.

When base and power both are the functions of x that is the function is of the form of $\sf [f(x)]^{g(x)}$

$\sf y = [f(x)]^{g(x)}$

$\sf log (y) = g(x).log[f(x)]$

$\sf \dfrac{1}{y} . \dfrac{dy}{dx} = \dfrac{d[g(x)] . log [ f(x)]}{dx}$

$\sf \dfrac{dy}{dx} = [f(x)]^{g(x)} \bigg[ \dfrac{d[g(x)] . log[f(x)]}{dx} \bigg]$

$\sf \dfrac{dy}{dx} = Q . \dfrac{d\bigg(power \times log (base) \bigg)}{dx}$

Answer 2

Answer:

Hope this helps u.. And correct too ✌️