Differentiation Formulas — Question 7

Question 7

Let $$f(x) = \frac{e^x \sin x}{x^2 + 1}$$

• Differentiate $f(x)$ using the quotient rule.

• Evaluate $f'(0)$.

Question 7 - Solution

We are given: $$f(x) = \frac{e^x \sin x}{x^2 + 1}$$

(a) Use the Quotient Rule:

The quotient rule is: $$f(x) = \frac{u(x)}{v(x)} \Rightarrow f'(x) = \frac{u'v - uv'}{v^2}$$

Let: $$u(x) = e^x \sin x, \quad v(x) = x^2 + 1$$

Differentiate $u(x)$ using the product rule: $$u'(x) = \frac{d}{dx}[e^x \sin x] = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$$

Differentiate $v(x)$: $$v'(x) = \frac{d}{dx}[x^2 + 1] = 2x$$

Now apply the quotient rule: $$f'(x) = \frac{e^x(\sin x + \cos x)(x^2 + 1) - e^x \sin x \cdot 2x}{(x^2 + 1)^2}$$

Factor out $e^x$: $$f'(x) = \frac{e^x \left[(\sin x + \cos x)(x^2 + 1) - 2x \sin x \right]}{(x^2 + 1)^2}$$

(b) Evaluate $f'(0)$:

Substitute $x = 0$:

Numerator: $$e^0 \left[ (\sin 0 + \cos 0)(0^2 + 1) - 2(0)\sin 0 \right] = 1 \cdot (0 + 1)(1) = 1$$

Denominator: $$(0^2 + 1)^2 = 1^2 = 1$$

So: $$f'(0) = \boxed{1}$$