Question

Differentiate with respect to X

not copied answer..


full solve it ..

3 questions

100 marks ✅✅​

Answer 1

ANSWER

1---

Logarithmic differentiation.

y = xsin x

ln y = (sin x) ln x ← taking the natural log of both sides.

y'/y = (cos x) ln x + (sin x)/x ← Using the product rule and implicit differentiation.

y' = y((cos x) ln x + (sin x)/x) ← multiplying by y.

y' = xsin x ((cos x) ln x + (sin x)/x) ← substitution.

Note that if you use logarithmic differentiation to differentiate y = uv with respect to x, you will get the following:

y' = vuv-1 * du/dx + uv ln u * dv/dx.

ANSWER

2--

Let us assume that the question as logx(here x=cosx).

Now,differentiation of logx gives 1/x.

So,it will become 1/cosx.

Again we should differentiate cosx which gives -sinx.

So,it will become -sinx/cosx=-tanx.

The final answer for diff of logcosx is -tanx.

Answer

3--

Assuming logxlog⁡x is base ee, let y=xlogxy=xlog⁡x. Take a logarithm of both sides to get logy=log(xlogx)=logxlogxlog⁡y=log⁡(xlog⁡x)=log⁡xlog⁡x. Differentiate both sides to get y′y=2logxxy′y=2log⁡xx Multiplying by yy gives y′=xlogx2logxx

Answer 2

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