by Rahul Anand | May 9, 2023 | CALCULUS
To prove : lim\( _{x \to a} { x^n – a^n \over x – a } = na^{n-1} \) where n is a rational number Proof: Let \( x = a + h \) Then as \(x \to a \), we have \(h \to 0 \) Now, \( \lim_{x \to a} { x^n – a^n \over x – a }...
by Rahul Anand | May 9, 2023 | CALCULUS
Proof : We have, lim\(_{θ\to 0} { \dfrac {\mathrm tan \mathrm θ}{ \mathrm θ} } \) = lim\(_{θ\to 0} { \dfrac {\mathrm \sin \mathrm θ} {\mathrm θ \mathrm \cos\mathrm θ} } \) \( \{∵ \tan\theta = \dfrac...
by Rahul Anand | May 9, 2023 | CALCULUS
As θ → 0, we have cosθ → 1 Proof : When θ = 0, We have, lim\(_{θ\to 0} \cos \)θ = cos0 = 1 { ∵ cos0 = 1 } Hence, lim\(_{θ\to 0} \cos \)θ = 1
by Rahul Anand | May 9, 2023 | CALCULUS
Let y = \(\mathsf {x^{n} }\) ∴ y + δy = \(\mathsf { {(x + δx)^{n}} }\) ∴ δy = y + δy – y = \(\mathsf { (x + δx)^{n} }\) – \(\mathsf { x^{n} }\) or δy = \(\mathsf { [\text{ }^{n}C_0 x^{n}{(δx)}^{0} }\) + \(\mathsf {\text{ }^{n}C_1 x^{n-1}{(δx)}^{1}}\) +...
by Rahul Anand | May 9, 2023 | CALCULUS
Derivative of \({e}^x\) using the First Principle Let \(y\) = \({e}^x\)∴ \(y + δy\) = \({e}^{x + δx}\)∴ \(δy\) = \({e}^{x + δx}\) – \({e}^x\)or \(δy\) = \({e}^{x}\) . \( [ {e}^{δx} – 1 ]\)Dividing each side by δx </h3>or \(\dfrac {δy}{δx}\) = \(...
by Rahul Anand | May 9, 2023 | CALCULUS
Derivative of \( sinθ \) using the First Principle Let \(y\) = \( sinθ \) ∴ \(y + δy\) = \( sin(θ + δθ) \) ∴ \(δy\) = \( sin(θ + δθ) \) – \( sinθ \)From Trigonometry , we have \( sin(A-B) \) = 2.\( sin \dfrac {(A-B)}{2} \).\( cos \dfrac {(A+B)}{2} \)Using the...
