Question 60 of 70intermediate🔧 ApplyShort Answer1 mark

Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers): a+bsin⁡xc+dcos⁡x\frac{a + b \sin x}{c + d \cos x}

Correct Answer

bccos⁡x+adsin⁡x+bd(c+dcos⁡x)2\frac{bc \cos x + ad \sin x + bd}{(c + d \cos x)^2}

Exercise: Miscellaneous Exercise on Chapter 12 | Q: 20 | (Page Page 37)
For More Understanding

Explanation

To find the derivative of the given function, we apply the Quotient Rule as demonstrated in the provided context for functions like cot⁡x=cos⁡xsin⁡x\cot x = \frac{\cos x}{\sin x}. We treat the numerator and denominator as separate functions uu and vv. The context confirms that the derivative of a constant is 0 and provides the trigonometric identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 for simplification.

Solution Steps

  1. Step 1: Let u=a+bsin⁡xu = a + b \sin x and v=c+dcos⁡xv = c + d \cos x. Differentiating with respect to xx, we get u′=bcos⁡xu' = b \cos x and v′=−dsin⁡xv' = -d \sin x.

  2. Step 2: Apply the Quotient Rule: ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.

  3. Step 3: Substitute the expressions for u,v,u′,v′u, v, u', v' into the formula: (bcos⁡x)(c+dcos⁡x)−(a+bsin⁡x)(−dsin⁡x)(c+dcos⁡x)2\frac{(b \cos x)(c + d \cos x) - (a + b \sin x)(-d \sin x)}{(c + d \cos x)^2}.

  4. Step 4: Simplify the numerator by expanding: bccos⁡x+bdcos⁡2x+adsin⁡x+bdsin⁡2xbc \cos x + bd \cos^2 x + ad \sin x + bd \sin^2 x.

  5. Step 5: Rearrange terms and use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1: bccos⁡x+adsin⁡x+bd(sin⁡2x+cos⁡2x)=bccos⁡x+adsin⁡x+bdbc \cos x + ad \sin x + bd(\sin^2 x + \cos^2 x) = bc \cos x + ad \sin x + bd.

  6. Step 6: The final derivative is bccos⁡x+adsin⁡x+bd(c+dcos⁡x)2\frac{bc \cos x + ad \sin x + bd}{(c + d \cos x)^2}.

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