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

If $x^2+y^2+\sin y=4$,then the value of $\frac{d^2 y}{d x^2}$ at $x=-2$ is

If $y=e^{ax}(\cos bx+\sin bx)$ satisfies the equation $\frac{d^2y}{dx^2}-K\frac{dy}{dx}+Ly=0$, then $L+bK=$

$A$ function $y = f(x)$ has a second-order derivative $f''(x) = 6(x - 1)$. If its graph passes through the point $(2, 1)$ and at that point the tangent to the graph is $y = 3x - 5$,then the function is

If $f(x) = f'(x) + f''(x) + f'''(x) + \ldots$ and $f(0) = 1$,then $f(x)$ is equal to

If $y=e^{\tan ^{-1} x}$, then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module