Question

differentiate x2 cosx from first principle..

Answer 1

Solution:

Differentiate x2 cosx from first principle:

➜ f(x) = cos (x² + 1) f(x + h) = cos (x + h) + 1)

➜ f'(x) = lim f(x + h) - f(x)

ₕ→₀ -----------------

h

➜ lim cos((x + h)² + 1) - cos(x² - 1)

ₕ→₀ ------------------------------------

h

Limit extending to 0:

➜ cos((x + 0)² + 1) - cos (x² + 1)

---------------------------------------

0

➜ cos(x² + 1) - cos (x² + 1) 0

------------------------------------ = ---- form.

0 0

L - hospital rule:

➜ – cos((x + 0)² + 1) [ 2 (n + 0) (1) + 0 ]

➜ – sin (x² + 1) [ 2x + 0 ]

➜ – sin (x² + 1) - 2x • sin [ x² + 1 ]

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Answer 2

differentiation of  x2 cosx from first principle.. is 2x cosx - x^2 sinx

• Given,
• f(x) = x^2 cosx

• First principle,
• $f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$
• $f'(x) = \lim_{h \to 0} \dfrac{(x+h)^2cos(x+h) - x^2cosx}{h}$
• Using L-Hospital's rule, we have,
• $f'(x) = \lim_{h \to 0} \dfrac{2(x+h)cos(x+h) + (x+h)^2[-sin(x+h)] -0}{1}$
• $f'(x) = \lim_{h \to 0} 2(x+h)cos(x+h) + (x+h)^2[-sin(x+h)]$
• $f'(x) = 2(x+0)cos(x+0) + (x+0)^2[-sin(x+0)]$
• $f'(x) = 2xcosx + x^2[-sinx]$
• ∴$f'(x) = 2xcosx - x^2sinx$