Learning these TS Inter 1st Year Maths 1B Formulas Chapter 9 Differentiation will help students to solve mathematical problems quickly.
TS Inter 1st Year Maths 1B Differentiation Formulas
→ Formula for finding derivative f'(x) of a function y = f(x) using the definition is
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = f'(x) = \({Lt}_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\)
Derivative of a function at a point ‘a’ f'(a) = \({Lt}_{x \rightarrow a}\left[\frac{f(x)-f(a)}{x-a}\right]\)
→ \(\frac{\mathrm{d}}{\mathrm{dx}}\)(u ± v) = \(\frac{d u}{d x} \pm \frac{d v}{d x}\)
→ \(\frac{d}{d x}\)(uv) = u.\(\frac{d v}{d x}\) + v.\(\frac{d u}{d x}\)
→ \(\frac{d}{d x}\) (uvw) = uv \(\frac{d}{d x}\)(w) + uw\(\frac{d}{d x}\)(v) + vw\(\frac{d}{d x}\)(u)
→ \(\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^2}\)
→ \(\frac{d}{d x}\)(xn) = n.xn-1
→ \(\frac{d}{d x}\left(\frac{1}{x^n}\right)=\frac{-n}{x^{n+1}}\)
→ \(\frac{d}{d x}\)(log x) = \(\frac{1}{x}\), \(\frac{d}{d x}\)(loga x) = loga e
→ \(\frac{d}{d x}\)(ex) = ex, \(\frac{d}{d x}\)(ax) = ax loge a
→ \(\frac{d}{d x}\)(sin x) = cos x
→ \(\frac{d}{d x}\)(cos x) = -sin x
→ \(\frac{d}{d x}\)(tan x) = sec2 x
→ \(\frac{d}{d x}\)(cot x) = -cosec2 x
→ \(\frac{d}{d x}\)(sec x) = sec x tan x
→ \(\frac{d}{d x}\)(cosec x) = -cosec x cot x
→ \(\frac{d}{d x}\)(sin-1x) = \(\frac{1}{\sqrt{1-x^2}}\)
→ \(\frac{d}{d x}\)(cos-1x) = \(-\frac{1}{\sqrt{1-x^2}}\)
→ \(\frac{d}{d x}\)(tan-1x) = \(\frac{1}{1+x^2}\)
→ \(\frac{d}{d x}\)(cot-1x) = \(-\frac{1}{1+x^2}\)
→ \(\frac{d}{d x}\)(sec-1x) = \(\frac{1}{|x| \sqrt{x^2-1}}\)
→ \(\frac{d}{d x}\)(cosec-1x) = \(-\frac{1}{|x| \sqrt{x^2-1}}\)
→ \(\frac{d}{d x}\)(sinh-1x) = \(\frac{1}{\sqrt{1+x^2}}\)
→ \(\frac{d}{d x}\)(cosh-1x) = \(\frac{1}{\sqrt{x^2-1}}\)
→ \(\frac{d}{d x}\)(tanh-1x) = \(\frac{-1}{1-x^2}\)
→ \(\frac{d}{d x}\)(coth-1x) = \(\frac{1}{1-x^2}\)
→ \(\frac{d}{d x}\)(sech-1x) = \(-\frac{1}{|x| \sqrt{1-x^2}}\)
→ \(\frac{d}{d x}\)(cosech-1x) = \(\frac{1}{|x| \sqrt{x^2+1}}\)
→ Logarithmic differentiation: If y = f(x)g(x) > then log y = g(x) log f(x)
⇒ \(\frac{1}{y} \frac{d y}{d x}\) = g(x).\(\frac{1}{f(x)}\)f'(x) + log[f(x)]g'(x)
→ Derivative of one function w.r.t. another function: If y = f(x); z = g(x) then \(\frac{d y}{d z}=\frac{d f}{d g}=\frac{f^{\prime}(x)}{g^{\prime}(x)}\), It is called as chain rule.
→ Parametric differentiation: If x = f(t), y = g(t) then \(\frac{d y}{d x}=\frac{d y}{\frac{d t}{d t}}=\frac{g^{\prime}(t)}{f^{\prime}(t)}\), Itis called as chain rule.
\(\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\left[\frac{d}{d t}\left(\frac{d y}{d x}\right)\right]\left(\frac{d t}{d x}\right)\)