Let fx be differentiable at x - h- then limx - h x - hf x - 2hf h x - h is equal to

The correct option is A f(h) + 2hf'(h)given that f(x) is diffeentiable at x=h, (x+h)f(x) 2h(f(h))x h is in the form of #160; 00 at h= 0 ∴lim _x→ ...

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Byju's Answer

Standard XII

Mathematics

First Principle of Differentiation

Let fx be dif...

Question

Let f(x) be differentiable at x = h, then $lim x \to h \frac{(x + h) f (x) - 2 h f (h)}{x - h}$ is equal to

f(h) + 2hf'(h)

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2f(h) + hf'(h)

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hf(h) + 2f'(h)

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hf(h) - 2f'(h)

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Solution

The correct option is A f(h) + 2hf'(h)
given that f(x) is diffeentiable at x=h,
$\frac{(x + h) f (x) - 2 h (f (h))}{x - h}$ is in the form of $\frac{0}{0}$ at h= 0
$∴ {lim}_{x \to h} \frac{(x + h) f (x) - 2 h (f (h))}{x - h} = l i m_{x \to h} \frac{\frac{(x + h) + f (x) - 2 h f (h)}{d x}}{\frac{d (x - h)}{d x}}$
$(∵ \frac{(d f (x)}{(d x} = f^{'} (x), h, f (x) a r e c o n s t a n t)$
$∴ {lim}_{x \to h} \frac{(x + h) f (x) - 2 h (f (h))}{x - h} = f (h) + 2 h f^{'} (h)$

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Let f(x) be differentiable at x = h, then limx→h(x+h)f(x)−2hf(h)x−h is equal to

The correct option is A f(h) + 2hf'(h)given that f(x) is diffeentiable at x=h,(x+h)f(x)−2h(f(h))x−h is in the form of 00 at h= 0∴limx→h(x+h)f(x)−2h(f(h))x−h=limx→h(x+h)+f(x)−2hf(h)dxd(x−h)dx(∵(df(x)(dx=f′(x),h,f(x)areconstant)∴limx→h(x+h)f(x)−2h(f(h))x−h=f(h)+2hf′(h)

Let f(x) be differentiable at x = h, then $lim x \to h \frac{(x + h) f (x) - 2 h f (h)}{x - h}$ is equal to