m is the slope of a tangent to the curve e^{y | vedclass

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

Question

(A) Given the curve equation: $e^{y} = 1 + x^2$. Taking the natural logarithm on both sides: $y = \ln(1 + x^2)$. To find the slope $m$,we differentiat

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the curve equation: $e^{y} = 1 + x^2$. Taking the natural logarithm on both sides: $y = \ln(1 + x^2)$. To find the slope $m$,we differentiate $y$ with respect to $x$: $m = \frac{dy}{dx} = \frac{d}{dx}(\ln(1 + x^2)) = \frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2)$. $m = \frac{1}{1 + x^2} \cdot (2x) = \frac{2x}{1 + x^2}$. Now,evaluate the slope at $x = 1$: $m = \frac{2(1)}{1 + (1)^2} = \frac{2}{1 + 1} = \frac{2}{2} = 1$. Thus,$m = 1$.

$m$ is the slope of a tangent to the curve $e^{y}=1+x^2$ at $x=1$,then $m=$

Similar Questions

If $\phi(x) = \log_{5} \log_{3} x$,then $\phi'(e)$ is equal to

If $f(x)=\log _{x^2}(\log x)$,then $f^{\prime}(x)$ at $x=e$ is

If $y = \log_3(\log_3 x)$,then $\frac{dy}{dx}$ at $x = 3$ is $\ldots \ldots$

Mrs. Rajni deposited $Rs. 10,000$ in a bank that pays $4 \%$ interest compounded continuously. The amount she gets after $10$ years is approximately $Rs.$ . . . . . . . (Given $e^{(0.4)} = 1.49182$)

$\frac{d}{dx} \left[ \log \left\{ e^x \left( \frac{x + 2}{x - 2} \right)^{3/4} \right\} \right]$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module