Chain Rule MCQ Quiz in தமிழ் - Objective Question with Answer for Chain Rule - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

Chain rule:

Let y = f(v) be a differentiable function of v and v = g(x) be a differentiable function of x then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} ⋅ \frac{{dv}}{{dx}}\)

Calculation:

Let y = cos(cos(x))

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}cos(cos(x))\)

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}= \ -\sin(cos(x))\frac{\mathrm{d} }{\mathrm{d} x}(cos(x))\)

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}=-sin(cos(x))(-sin(x))\)

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}=sin(cos(x))(sin(x))\)

Formula used:

\(\frac{d}{d x}(log x) =\frac{1}{x}\)

\(\frac{d}{d x}(\cos x) = - sin x\)

Calculation:

Let,

\(\frac{d}{d x}(\log \cos x)\)

(1/cos x) \(\frac{d}{d x}( \cos x)\)

(1/cos x) × (-  sin x)

∴ \(\frac{d}{d x}(\log \cos x)\) = - tan x

Given:

f(x) = cos(sin(x))

Concept used:

The chain rule states that if you have a composition of functions, like f(g(x)), the derivative is f′(g(x)) × g′(x).

Calculation:

⇒ f(u) = cos(u)

⇒ f'(u) = -sin(u) (derivative of cos(u))

⇒ g(x) = sin(x)

⇒ g'(x) = cos(x) (derivative of sin(x))

⇒ \(\dfrac{d}{dx}\)[cos(sin(x))] = f'(g(x)) × g'(x)

⇒ \(\dfrac{d}{dx}\)[cos(sin(x))] = -sin(sin(x)) × cos(x)

∴ The derivative of cos(sin(x)) with respect to x is −cos x ⋅ sin (sin x)

Explanation:

If f is a function of x only, then we revert back to the single-variable chain rule:

df/dx = df/dt × dt/dx

In this scenario, f depends on x indirectly through an intermediate variable t.

This expression is saying the derivative of f with respect to x is equal to the derivative of f with respect to t multiplied by the derivative of t with respect to x.

Concept:

Chain rule: \(\rm\frac{d}{d x}[f(g(x))]=f^{\prime}(g(x)) g^{\prime}(x)\)

Differentiation Formulas

\(\rm \frac{d\: (tan x)}{dx}\) = sec² x 

\(\rm \frac{d}{dx} (e^{x}) = e^{x} \)

\(\rm \frac{d}{dx} (x^{n}) = n x^{n - 1} \)​

Calculation:

Given: \(\rm \frac{d}{dx} (e^{\tan {x^2}}) \)

= \(\rm e^{\tan {x^2}} \) \(\rm \frac{d}{dx} (\tan x^2)\)

= \(\rm e^{\tan {x^2}} \)⋅sec² x²

= 2x⋅\(\rm e^{\tan {x^2}} \)⋅sec² x²

Concept:

Chain rule:

Let y = f(v) be a differentiable function of v and v = g(x) be a differentiable function of x then \(\frac{{dy}}{{dx}} = \frac{{dy}}{{dv}} ⋅ \frac{{dv}}{{dx}}\)

Calculation:

Let y = sin(sin(x))

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}sin(sin(x))\)

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}=cos(sin(x))\frac{\mathrm{d} }{\mathrm{d} x}(sin(x))\)

⇒ \(\rm \frac{\mathrm{d} y}{\mathrm{d} x}=cos(sin(x))(cos(x))\)

Concept:

Chain rule: \(\rm\frac{d}{d x}[f(g(x))]=f^{\prime}(g(x)) g^{\prime}(x)\)

\(\rm\frac{d}{d x}[\sin x]=cos x\)

\(\rm\frac{d}{d x}[\ e^x]=e^x\)

\(\rm\frac{d}{d x}[\log x]= {1\over x}\)

Calculation:

Given: f(x) =  log (sin e^x)

f'(x) = \(\rm 1\over {\sin e^x}\) ⋅ \(\rm\frac{d}{d x}[\sin e^x]\)

= \(\rm 1\over {\sin e^x}\)⋅ (cos e^x) ⋅ \(\rm\frac{d}{d x}(e^x)\)

= \(\rm {{\cos e^x}\over {\sin e^x}} ⋅ e^x\)              (∵ \(\rm {{\cos x}\over {\sin x}} = cot\ x\))

= e^x⋅ cot e^x

Explanation:

Given:

f(0) = -1 and f'(0) = 1 Let h(x) = f(2f(x) +2) and g(x) = (h(x))2 

⇒ g(x) = (h(x))²

⇒ g'(x) = 2(h(x)).h'(x)

⇒ g'(0) = 2(h(0)).h'(0)

= 2f(2f (0) + 2).2

= 2f(0).2 = (–2).2 = –4

∴ Option (a) is correct

Explanation:

Let f: (–1, 1)→ R be a differentiable function with f(0) = –1 and f'(0) = 1

⇒ h(x) = f(2f(x) + 2)

⇒ h'(x) = f'(2f(x) + 2)2f'(x)

⇒ h'(0) = f'(2f(0) + 2).2f'(0)

= f'(–2 + 2).2(1)

= f'(0).2 = (1).2 = 2

∴ Option (d) is correct.

Given:

Differentiate sin(x² + 9) with respect to x.

Formula used:

Chain Rule: If y = sin(u) and u = x² + 9, then dy/dx = cos(u) × du/dx

Calculation:

Let y = sin(x² + 9)

⇒ dy/dx = cos(x² + 9) × (d/dx)(x² + 9)

⇒ dy/dx = cos(x² + 9) × 2x

⇒ dy/dx = 2x cos(x² + 9)

∴ The correct answer is option (4).