MHT-CET Differentiation — practice questions
30 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice MHT-CET Differentiation in the app →$\dfrac{d}{dx}(x^n)$ equals:$\dfrac{d}{dx}( in x)$ equals:$\dfrac{d}{dx}(e^x)$ equals:$\dfrac{d}{dx}(\ln x)$ equals:The **chain rule** states $\dfrac{dy}{dx}$ for $y = f(g(x))$ is:The product rule for $\dfrac{d}{dx}(uv)$ is:$\dfrac{d}{dx}(\tan x)$ equals:$\dfrac{d}{dx}( in^{-1} x)$ equals:If $y = in(\cos x)$, then $\dfrac{dy}{dx}$ equals:If $y = x^x$, then $\dfrac{dy}{dx}$ equals:If $y = \log( in x)$, then $\dfrac{dy}{dx}$ equals:$\dfrac{d}{dx}(a^x)$ equals (where $a > 0$):If $x^2 + y^2 = 25$, then $\dfrac{dy}{dx}$ equals:If $x = a t^2$, $y = 2 a t$ (parametric), then $\dfrac{dy}{dx}$ equals:If $y = in x$, then $y_n = \dfrac{d^n y}{dx^n}$ at $x = 0$ for $n = 4$ equals:If $y = \tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$, then $\dfrac{dy}{dx}$ equals:If $y = \tan^{-1}\left( qrt{\dfrac{1-\cos x}{1+\cos x}}\right)$, then $\dfrac{dy}{dx}$ equals:If $y = x in x$, then $y_2 = \dfrac{d^2y}{dx^2}$ equals:If $x^y = y^x$, then $\dfrac{dy}{dx}$ equals:If $y = in^{-1}(2x qrt{1-x^2})$ for $-1/ qrt 2 < x < 1/ qrt 2$, then $\dfrac{dy}{dx}$ equals:$\dfrac{d}{dx}\left[\log(\log x)\right]$ at $x = e^e$ equals:If $y = ( in x)^x$, then $\dfrac{dy}{dx}$ equals:Given $x = a(\theta - in\theta)$, $y = a(1 - \cos\theta)$ (cycloid). Then $\dfrac{dy}{dx}$ equals:If $y = e^x \cos x$, then $\dfrac{d^2 y}{dx^2} - 2\dfrac{dy}{dx} + 2y$ equals:If $f(x) = |x - 1| + |x - 3|$, then $f'(2)$ equals:If $y = \tan^{-1}\left(\dfrac{ qrt{1+x^2} - 1}{x}\right)$, then $\dfrac{dy}{dx}$ equals:If $y_n$ denotes the $n$-th derivative of $y = in x$, then $y_n = $:If $f(x) = x^3 - 3x$, the values of $x$ where $f'(x) = 0$ are:If $f(x) = e^{ax} in(bx)$, then $f''(x) + a^2 f(x)$ equals:$\dfrac{d}{dx}\left[\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right)\right]$ for $x > 0$ equals: