Find frac{d^{2} y}{d x^{2}}, if y=x^{3}+tan x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $y = x^{3} + 	an x$. First,we find the first derivative with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^{3}) + \frac{d}{dx}(	an

Vedclass · Accepted Answer

$6 x + 2 \sec^{2} x 	an x$

Vedclass · Answer

(A) Given that $y = x^{3} + 	an x$. First,we find the first derivative with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^{3}) + \frac{d}{dx}(	an x) = 3x^{2} + \sec^{2} x$. Now,we find the second derivative by differentiating $\frac{dy}{dx}$ with respect to $x$: $\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(3x^{2} + \sec^{2} x) = \frac{d}{dx}(3x^{2}) + \frac{d}{dx}(\sec^{2} x)$. Using the power rule and chain rule: $\frac{d}{dx}(3x^{2}) = 6x$. For $\frac{d}{dx}(\sec^{2} x)$,let $u = \sec x$,then $\frac{d}{dx}(u^{2}) = 2u \frac{du}{dx} = 2(\sec x)(\sec x 	an x) = 2 \sec^{2} x 	an x$. Thus,$\frac{d^{2}y}{dx^{2}} = 6x + 2 \sec^{2} x 	an x$.

Find $\frac{d^{2} y}{d x^{2}},$ if $y=x^{3}+\tan x$.

Similar Questions

There exists a function $f(x)$ satisfying $f(0) = 1$,$f'(0) = -1$,and $f(x) > 0$ for all $x$. Which of the following must be true for $f''(x)$?

If $y = \cos^{-1} x$,find $\frac{d^{2} y}{d x^{2}}$ in terms of $y$ alone.

If ${y^2} = a{x^2} + bx + c$,then ${y^3}\frac{{{d^2}y}}{{d{x^2}}}$ is

If $e^{y}(x+1)=1,$ show that $\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}$.

Let $f : S \rightarrow S$ where $S =(0, \infty)$ be a twice differentiable function such that $f(x+1) = xf(x)$. If $g : S \rightarrow R$ is defined as $g(x) = \log_{e} f(x)$,then the value of $|g''(5) - g''(1)|$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module