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NCERT Exemplar · Q72

Q.If y=sin⁡(x+9)cos⁡xy = \dfrac{\sin(x + 9)}{\cos x} then dydx\dfrac{dy}{dx} at x=0x = 0 is
(A) cos⁡9\cos 9
(B) sin⁡9\sin 9
(C) 00
(D) 11

Assam AhsecMCQ· 1mImportance★★★★★est
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To find the derivative of the given quotient function at x=0x=0, we first apply the Quotient Rule, simplify the resulting expression using a trigonometric identity, and then substitute x=0x=0. The final value is cos⁡9\boxed{\cos 9}.

When we are asked to find dydx\frac{dy}{dx} at a specific point, say x=0x=0, it means we need to determine the instantaneous rate of change of the function yy with respect to xx precisely at that point. Geometrically, this represents the slope of the tangent line to the curve y=f(x)y=f(x) at x=0x=0.

The general approach involves two main steps:

  1. Find the general derivative dydx\frac{dy}{dx}: This will be an expression in terms of xx.
  2. Evaluate the derivative at the given point: Substitute x=0x=0 into the expression obtained in step 1.

Our function y=sin⁡(x+9)cos⁡xy = \dfrac{\sin(x + 9)}{\cos x} is a quotient of two functions of xx. Therefore, the most direct way to find its derivative is by using the Quotient Rule.

If y=uvy = \frac{u}{v}, where uu and vv are differentiable functions of xx, then dydx=vdudx−udvdxv2\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}.

Let's apply this rule step-by-step.

  1. Identify uu and vv and their derivatives.

    Let u=sin⁡(x+9)u = \sin(x+9) and v=cos⁡xv = \cos x.

    Now, we find their derivatives with respect to xx:

    • For u=sin⁡(x+9)u = \sin(x+9), we use the chain rule. Let z=x+9z = x+9, so u=sin⁡zu = \sin z. Then dudx=dudz⋅dzdx\frac{du}{dx} = \frac{du}{dz} \cdot \frac{dz}{dx}. dudz=cos⁡z=cos⁡(x+9)\frac{du}{dz} = \cos z = \cos(x+9). dzdx=ddx(x+9)=1\frac{dz}{dx} = \frac{d}{dx}(x+9) = 1. So, dudx=cos⁡(x+9)⋅1=cos⁡(x+9)\frac{du}{dx} = \cos(x+9) \cdot 1 = \cos(x+9).
    • For v=cos⁡xv = \cos x, its derivative is dvdx=−sin⁡x\frac{dv}{dx} = -\sin x.
  2. Apply the Quotient Rule.

    Substitute u,v,dudx,dvdxu, v, \frac{du}{dx}, \frac{dv}{dx} into the Quotient Rule formula:

dydx=cos⁡x⋅cos⁡(x+9)−sin⁡(x+9)⋅(−sin⁡x)(cos⁡x)2\frac{dy}{dx} = \frac{\cos x \cdot \cos(x+9) - \sin(x+9) \cdot (-\sin x)}{(\cos x)^2}

dydx=cos⁡xcos⁡(x+9)+sin⁡xsin⁡(x+9)cos⁡2x\frac{dy}{dx} = \frac{\cos x \cos(x+9) + \sin x \sin(x+9)}{\cos^2 x}

  1. Simplify the expression using a trigonometric identity. The numerator, cos⁡xcos⁡(x+9)+sin⁡xsin⁡(x+9)\cos x \cos(x+9) + \sin x \sin(x+9), is a standard trigonometric identity for the cosine of a difference of two angles: cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A-B) = \cos A \cos B + \sin A \sin B. Here, A=xA=x and B=x+9B=x+9. So, the numerator simplifies to cos⁡(x−(x+9))=cos⁡(x−x−9)=cos⁡(−9)\cos(x - (x+9)) = \cos(x - x - 9) = \cos(-9). …

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