# Tutor profile: Nidhi J.

## Questions

### Subject: Pre-Calculus

Check whether $$e^x$$ has a maxima or minima.

Let $$f(x)=e^x$$ Then, $$f'(x)= e^x$$ Now, if $$f'(x)= 0$$, then $$e^x=0$$ But, the exponential function can never assume 0 for any x. So, there does not exist any real x such that $$f'(x)= 0$$. Hence, $$e^x$$ does not have any maxima or minima.

### Subject: Calculus

If $$x^y=e^{x-y}$$, then find $$\frac{dy}{dx}$$.

Taking $$log_e$$ on both sides of the equation, we get $$log_e{x^y}=log_ee^{x-y}$$ (Using $$loga^b=bloga$$ and $$log_ee=1$$) $$ylog_ex={(x-y)}log_ee$$ $$ylog_ex=x-y$$ $$ylog_ex+y=x$$ $$y(log_ex+1)=x$$ $$y=\frac{x}{(log_ex+1)}$$ Now, $$\frac{dy}{dx}= \frac{(1+log_ex).1-x{(\frac{1}{x}})}{(1+log_ex)^2}$$ $$\frac{dy}{dx}= \frac{1+log_ex-1}{(1+log_ex)^2}$$ $$\frac{dy}{dx}= \frac{log_ex}{(1+log_ex)^2}$$

### Subject: Statistics

10% of toys produced in a certain manufacturing process were found defective. Find the probability that in a sample of 10 toys chosen at random, exactly 2 will be defective by using binomial distribution.

Since there are only 2 outcomes (defective or non-defective) for each independent trial, so we can use Binomial distribution here. Take, $$n=10$$, $$x=2$$ First we find probability of defective toy, $$p=10 \% =\frac{10} {100}=0.1$$ So, probability of non-defective toy, $$q=1-p=0.9$$ By Binomial distribution, Probability of 2 defective toy $$=^nC_xp^xq^{n-x} $$ $$=^{10}C_2{(0.1)}^2{(0.9)^8}$$ $$=\frac{10!}{2!8!}{(0.1)}^2{(0.9)}^8$$ $$=\frac{45}{{10}^{10}}.9^8$$ $$=0.1937$$

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