Question

what is the derivative of logsinx​

Answer 1

Answer:

Hope it help you:)

Step-by-step explanation:

In order to answer this question, we have to first understand how to differentiate log(x)log(x)is, since log(sin(x))log(sin(x)) is differentiated similarly to this.

Consider y=loga(x)y=loga(x), if we wanted to express this equation in exponential form, we would raise both sides of the equation by the base aa, resulting in our new equation: ay=xay=x. Now if we were to solve for y, we could either take the log of both sides, or the natural log if we wanted to in order to turn y into a coefficient as opposed to an exponent and solve for it. Since we are more familiar with differentiating the natural log (ddxln(x)=1x)(ddxln(x)=1x), we will take the natural log of both sides: ln(ay)=ln(x)ln(ay)=ln(x)→yln(a)=ln(x)yln(a)=ln(x)→y=ln(x)ln(a)y=ln(x)ln(a). Therefore y=loga(x)=ln(x)ln(a)y=loga(x)=ln(x)ln(a), which means ddxloga(x)=ddxln(x)ln(a)ddxloga(x)=ddxln(x)ln(a). Applying this to information in differentiating log(x)log(x) leads to: ddxlo

Answer 2

Hint – First principle of derivatives says that for f(x), f,(x)=limh→0f(x+h)−f(x)h. So, apply the given equation to the function given in the question.

Complete step-by-step answer:

Let f(x)=sinx

First principle of derivatives says that for f(x),

f,(x)=limh→0f(x+h)−f(x)h.

So, we shall apply it to the function given.

f(x)=logsinx

f,(x)=limh→0log(sin(x+h))−log(sinx)h byfirstprinciple

Using the addition formula, we get

sin(A+B)=sinAcosB+cosAsinB.

So, the above equation will transform into-

f,(x)=limh→0log(sin(x)cos(h)+cos(x)sin(h))−log(sinx)h

We can also use the subtraction that says,

loga(b)−loga(c)=loga(bc) to get, so, now the equation will be-

f,(x)=limh→0log(sin(x)cos(h)+cos(x)sin(h)sinx)h

Now, dividing numerator by sin x we get-

=limh→0log(cos(h)+cot(x)sin(h))h

Now, cos (h) = 1, as h tends to 0.

Therefore, the equation becomes-

=limh→0log(1+cot(x)sin(h))h

Divide and multiply by cot(x).sin(h) , we get-

=limh→0log(1+cot(x)sin(h))h(cot(x).sin(h)).cot(x)sin(h)

Now using the property log(1+xx)=1, so this implies that-

log(1+cotx.sinh)cotx.sinh=1 .

Therefore, the above equation becomes-

=limh→01h.cot(x)sin(h)

Now we know that limh→01hsin(h)=1

. So, the equation becomes now-

=cot(x)

Hence, the derivative of logsinx by first principle is cot (x).

Note- Whenever such types of question appear then always proceed using the formula f,(x)=limh→0f(x+h)−f(x)h and be careful about evaluating limits. Just make sure that you didn’t skip any step as it is a long solution. Make the necessary assumptions when needed.