Question

Question:-


if y=x^tan x +(sin x)^cos x, find dy/dx​​

Answer 1

Given:-

$\bf y=x^{tanx}+(sinx)^{cosx}$

To Find:-

$\bf\dfrac{dy}{dx}$

Solution:-

Apply ln on both sides ,

$\begin{gathered}\implies \sf ln(y)=ln(x^{tanx})+ln(sinx^{cosx})\\\\\implies \sf ln(y)=tanx.lnx+cosx.ln(sinx)\;[\;Since\ ln(a^b)=b.lna]\end{gathered}$

Now , differentiate y with respect to x ,

$\begin{gathered}\implies \sf \dfrac{d}{dx}(ln(y))=\dfrac{d}{dx}(tanx.lnx+cosx.ln(sinx))\\\\\implies \sf \dfrac{1}{y}.\dfrac{dy}{dx}=\dfrac{d}{dx}(tanx.lnx)+\dfrac{d}{dx}(cosx.ln(sinx))\;[\;Since\;,\dfrac{d}{dx}(lnx)=\dfrac{1}{x\;}]\\\\\implies \sf \dfrac{1}{y}.\dfrac{dy}{dx}=lnx.\dfrac{d}{dx}(tanx)+tanx\dfrac{d}{dx}(lnx)+ln(sinx)\dfrac{d}{dx}(cosx)+cosx\dfrac{d}{dx}(ln(sinx))\end{gathered}$

$\begin{gathered}\sf Since\;,\dfrac{d}{dx}(uv)=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\\\\\implies \sf \dfrac{1}{y}.\dfrac{dy}{dx}=lnx.sec^2x+\dfrac{tanx}{x}+ln(sinx)(-sinx)+cosx.\dfrac{1}{sinx}.cosx\\\\\implies \bf \dfrac{dy}{dx}=y(lnx.sec^2x+\dfrac{tanx}{x}-sinx.ln(sinx)+cotx.cosx)\\\\\implies \sf \dfrac{dy}{dx}=(x^{tanx}+(sinx)^{cosx})(lnx.sec^2x+\dfrac{tanx}{x}-sinx.ln(sinx)+cotx.cosx) \end{gathered}$

Answer:-

$\bf \dfrac{dy}{dx}=(x^{tanx}+(sinx)^{cosx})(lnx.sec^2x+\dfrac{tanx}{x}-sinx.ln(sinx)+cotx.cosx)$

Hope It's Helps Uh :D

Answer 2

Answer:

Given:-

\bf y=x^{tanx}+(sinx)^{cosx}y=x

tanx

+(sinx)

cosx

To Find:-

\bf\dfrac{dy}{dx}

dx

dy

Solution:-

Apply ln on both sides ,

\begin{gathered}\begin{gathered}\implies \sf ln(y)=ln(x^{tanx})+ln(sinx^{cosx})\\\\\implies \sf ln(y)=tanx.lnx+cosx.ln(sinx)\;[\;Since\ ln(a^b)=b.lna]\end{gathered}\end{gathered}

⟹ln(y)=ln(x

tanx

)+ln(sinx

cosx

)

⟹ln(y)=tanx.lnx+cosx.ln(sinx)[Since ln(a

b

)=b.lna]

Now , differentiate y with respect to x ,

\begin{gathered}\begin{gathered}\implies \sf \dfrac{d}{dx}(ln(y))=\dfrac{d}{dx}(tanx.lnx+cosx.ln(sinx))\\\\\implies \sf \dfrac{1}{y}.\dfrac{dy}{dx}=\dfrac{d}{dx}(tanx.lnx)+\dfrac{d}{dx}(cosx.ln(sinx))\;[\;Since\;,\dfrac{d}{dx}(lnx)=\dfrac{1}{x\;}]\\\\\implies \sf \dfrac{1}{y}.\dfrac{dy}{dx}=lnx.\dfrac{d}{dx}(tanx)+tanx\dfrac{d}{dx}(lnx)+ln(sinx)\dfrac{d}{dx}(cosx)+cosx\dfrac{d}{dx}(ln(sinx))\end{gathered}\end{gathered}

⟹

dx

d

(ln(y))=

dx

d

(tanx.lnx+cosx.ln(sinx))

⟹

y

1

.

dx

dy

=

dx

d

(tanx.lnx)+

dx

d

(cosx.ln(sinx))[Since,

dx

d

(lnx)=

x

1

]

⟹

y

1

.

dx

dy

=lnx.

dx

d

(tanx)+tanx

dx

d

(lnx)+ln(sinx)

dx

d

(cosx)+cosx

dx

d

(ln(sinx))

\begin{gathered}\begin{gathered}\sf Since\;,\dfrac{d}{dx}(uv)=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\\\\\implies \sf \dfrac{1}{y}.\dfrac{dy}{dx}=lnx.sec^2x+\dfrac{tanx}{x}+ln(sinx)(-sinx)+cosx.\dfrac{1}{sinx}.cosx\\\\\implies \bf \dfrac{dy}{dx}=y(lnx.sec^2x+\dfrac{tanx}{x}-sinx.ln(sinx)+cotx.cosx)\\\\\implies \sf \dfrac{dy}{dx}=(x^{tanx}+(sinx)^{cosx})(lnx.sec^2x+\dfrac{tanx}{x}-sinx.ln(sinx)+cotx.cosx) \end{gathered}\end{gathered}

Since,

dx

d

(uv)=v

dx

du

+u

dx

dv

⟹

y

1

.

dx

dy

=lnx.sec

2

x+

x

tanx

+ln(sinx)(−sinx)+cosx.

sinx

1

.cosx

⟹

dx

dy

=y(lnx.sec

2

x+

x

tanx

−sinx.ln(sinx)+cotx.cosx)

⟹

dx

dy

=(x

tanx

+(sinx)

cosx

)(lnx.sec

2

x+

x

tanx

−sinx.ln(sinx)+cotx.cosx)

Answer:-

\bf \dfrac{dy}{dx}=(x^{tanx}+(sinx)^{cosx})(lnx.sec^2x+\dfrac{tanx}{x}-sinx.ln(sinx)+cotx.cosx)

dx

dy

=(x

tanx

+(sinx)

cosx

)(lnx.sec

2

x+

x

tanx

−sinx.ln(sinx)+cotx.cosx)

Hope It's Helps Uh :D