Question

Find dy/dx when y=sinx/x^3

Question of differentiation, IIT

Answer 1

Answer:

we will apply the quotient rule

= cosx × x^3 - sinx × 3x^2/x^6

Answer 2

The answer is $\frac{cosx}{x^{3} } - \frac{3sinx}{x^{4} }$.

Explanation:

According to the given information, we need to find the the differentiation of y=sinx/x^3, that is, we need to find $\frac{dy}{dx}= \frac{d}{dx}(\frac{sinx}{x^{3}} )$.

Now, we know the rule of differentiation, that, $\frac{d}{dx} (\frac{a}{b})$, where a and b are functions of x and b is a non-zero number, is equal to $\frac{b\frac{d}{dx}(a) - a\frac{d}{dx}(b) }{b^{2} }$

Then, applying the similar formula to evaluate the problem, we get,

$\frac{x^{3} \frac{d}{dx}(sinx) - sinx\frac{d}{dx}(x^{3} ) }{(x^{3})^{2} }$...(1)

Now, the $\frac{d}{dx} sinx = cosx$

Also, $\frac{d}{dx} x^{3} = 3x^{2}$

Then, (1) becomes,

$\frac{x^{3} cosx - sinx(3x^{2} ) }{x^{6} }$

Simplifying, we get,

$\frac{cosx}{x^{3} } - \frac{3sinx}{x^{4} }$.

Hence, the answer is $\frac{cosx}{x^{3} } - \frac{3sinx}{x^{4} }$.

Learn more here

https://brainly.in/question/21825413

Learn more

https://brainly.in/question/8960809