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To find the derivative of xsinx, we take the limiting value as x approaches x + h. To simplify this, we set x = x + h, and we want to take the limiting value as h approaches 0. We will use the following formulas to prove the result: d (xsinx)/dx = lim h→0 [ (x+h)sin (x+h) - xsinx]/ [ (x+h) - x] = lim h→0 [x sin (x+h) + h sin (x+h) - x sinx]/h.

Replace y' y ′ with dy dx d y d x. dy dx = xcos(x)+sin(x) d y d x = x cos (x) + sin (x) Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with step-by-step explanations, just like a math tutor.

14 hours ago · In this case, k(x) = sin x, so the derivative of g(x) is: > d/dx [ln(sin x)] = 1/sin x * d/dx [sin x] = 1/sin x * cos x = cot x. The derivative of h(x) is given by: > d/dx [cos x] = -sin x. Now, we can plug these derivatives back into the quotient rule formula:

\(\dfrac{d}{dx}\big(\sin x\big)=\cos x\quad\text{and}\quad\dfrac{d}{dx}\big(\cos x\big)=−\sin x\). With these two formulas, we can determine the derivatives of all six basic trigonometric functions.

All derivatives of circular trigonometric functions can be found from those of sin (x) and cos (x) by means of the quotient rule applied to functions such as tan (x) = sin (x)/cos (x). Knowing these derivatives, the derivatives of the inverse trigonometric …

Dec 16, 2024 · Ex 5.5, 9 Differentiate the functions in, 𝑥^sin⁡𝑥 + 〖(sin⁡𝑥)〗^cos⁡𝑥 Let y = 𝑥^sin⁡𝑥 + 〖(sin⁡𝑥)〗^cos⁡〖𝑥 〗 Let 𝑢 =𝑥^sin⁡𝑥 & 𝑣 =〖(sin⁡𝑥)〗^cos⁡𝑥 ∴ 𝑦 = 𝑢 + 𝑣 Differentiating both sides 𝑤.𝑟.𝑡.𝑥.

table d^n/dx^n (sinx) for n = 1 ... 5; sin(x), sin'(x), sin''(x), sin'''(x), sin''''(x) d/dx {sinx, cosx, tanx, cotx, secx, cscx}

$$\dfrac{d}{dx} [\sin x] = \displaystyle{\lim_{h \rightarrow 0} \dfrac{\sin (x+h) - \sin (x)}{h}}$$ Recalling the trigonometric identity $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$, we can replace the $\sin(x + h)$ above to arrive at $$\dfrac{d}{dx} [\sin x] = \displaystyle{\lim_{h \rightarrow 0} \dfrac{\sin x \cos ...

Half-Angle Formulas. Applications: Rutherford scattering