10^{-x tan x} left[ frac{d}{dx} (10^{x tan x} | vedclass

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

Question

(B) Let $y = 10^{x 	an x}$.Using the chain rule for differentiation,$\frac{d}{dx}(a^u) = a^u \ln a \cdot \frac{du}{dx}$.Here,$a = 10$ and $u = x 	an

Vedclass · Accepted Answer

$\ln 10 (	an x + x \sec^2 x)$

Vedclass · Answer

(B) Let $y = 10^{x 	an x}$.Using the chain rule for differentiation,$\frac{d}{dx}(a^u) = a^u \ln a \cdot \frac{du}{dx}$.Here,$a = 10$ and $u = x 	an x$.First,find the derivative of $u = x 	an x$ using the product rule:$\frac{d}{dx}(x 	an x) = \frac{d}{dx}(x) \cdot 	an x + x \cdot \frac{d}{dx}(	an x) = 1 \cdot 	an x + x \cdot \sec^2 x = 	an x + x \sec^2 x$.Now,differentiate $10^{x 	an x}$:$\frac{d}{dx}(10^{x 	an x}) = 10^{x 	an x} \cdot \ln 10 \cdot (	an x + x \sec^2 x)$.Substitute this back into the original expression:$10^{-x 	an x} \left[ 10^{x 	an x} \cdot \ln 10 \cdot (	an x + x \sec^2 x) ight]$$= 10^{-x 	an x + x 	an x} \cdot \ln 10 \cdot (	an x + x \sec^2 x)$$= 10^0 \cdot \ln 10 \cdot (	an x + x \sec^2 x)$$= \ln 10 (	an x + x \sec^2 x)$.

$10^{-x \tan x} \left[ \frac{d}{dx} (10^{x \tan x}) \right]$ is equal to

Similar Questions

$\frac{d}{dx} \log_{\sqrt{x}} \left(\frac{1}{x}\right)$ is equal to

Let $f:(-1,1) \rightarrow \mathbb{R}$ be a differentiable function with $f(0)=-1$ and $f^{\prime}(0)=1$. If $g(x)=(f(2f(x)+2))^2$, then $g^{\prime}(0)=$

Let $f(x)$ be a differentiable function for all $x \in R$ and $f(x+y)=f(x)+f(y)-3xy$. If $\lim _{h \rightarrow 0} \frac{f(h)}{h}=7$, then $f^{\prime}(x)=$

Derivative of $e^x$ with respect to $\sqrt{x}$ is

If $f(x) = x \tan^{-1} x$,then $f'(1) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module