Question

#Quality Question

@ Derivatives

If
$\bf y = f( {x}^{3} ) \\ \\ \bf z = g( {x}^{5} ) \\ \\ \bf f(x) = tan \: x$

Then Find the value of
$\bf \huge \frac{dy}{dz}$
​

Answer 1

Given:

$\bf y = f( {x}^{3} ) \\ \\ \bf z = g( {x}^{5} ) \\ \\ \bf f'(x) = tan \: x \\ \\ \bf g'(x) = sec \: x$

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To Find:

$\huge \sf \frac{dy}{dz}$

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Solution:

$\sf y = f({x}^{3}) \\ \\ \sf\implies \frac{dy}{dx} = f'( {x}^{3} ).3 {x}^{2} \\ \\ \implies \underline{\boxed{ \bf \frac{dy}{dx} = 3 {x}^{2} .tan {x}^{3}} } \\ \{\because f' (x ) = tan \: x \}$

Also

$\sf z = g(x {}^{5} ) \\ \\ \sf \implies \frac{dz}{dx} = g'( {x}^{5} ).5 {x}^{4} \\ \\ \implies \bf \underline{ \boxed{ \bf\frac{dz}{dx} = 5 {x}^{4} .sec \: {x}^{5 }}}\\ \{ \because g'(x) = sec \: x \}$

Now,

$\sf \dfrac{dy}{dz} = \dfrac{ \frac{dy}{dx} }{ \frac{dz}{dx} } \\ \\ = \sf \dfrac{3 {x}^{2} \: tan x {}^{3} }{5 {x}^{4} \: secx {}^{5} } \\ \\ \large \red{ \implies \underline{ \boxed{\bf \frac{dy}{dz} = \dfrac{3}{5 {x}^{2} } \times \frac{tan {x}^{3} }{sec {x}^{5} } }}}$

$\large\mathfrak{ \text{W}hich \: \: is \: \: the \: \: required \: \: \text{ A}nswer.}$

Answer 2

We Have

$y = f({x}^{3}) \\ \\ \frac{dy}{dx} = f'( {x}^{3} ).3 {x}^{2} \\ \\ \boxed{\boxed{ \bf \frac{dy}{dx} = 3 {x}^{2} .tan {x}^{3}} }\:\: (GIVEN)$

$z = g(x {}^{5} ) \\ \\ \frac{dz}{dx} = g'( {x}^{5} ).5 {x}^{4} \\ \\ \bf \boxed{ \boxed{ \bf\frac{dz}{dx} = 5 {x}^{4} .sec \: {x}^{5 }}}\:\:(GIVEN)$

Now,

$\dfrac{dy}{dz} = \dfrac{ dy/dx}{ dz/dx } \\ \\ = \dfrac{3 {x}^{2} \: tan x {}^{3} }{5 {x}^{4} \: secx {}^{5} } \\ \\ \large { \colorbox{orange}{ \boxed{\bf \frac{dy}{dz} = \dfrac{3}{5 {x}^{2} } • \frac{tan {x}^{3} }{sec {x}^{5} } }}}$

So this is The final Answer to this question.